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Page No 16.10:

Question 1:

Find the slopes of the tangent and the normal to the following curves at the indicted points:

(i) y=x3 at x=4
(ii) y=x at x=9
(iii) y = x3x at x = 2
(iv) y = 2x2 + 3 sin x at x = 0
(v) x = a (θ − sin θ), y = a(1 − cos θ) at θ = −π/2
(vi) x = a cos3 θ, y = a sin3 θ at θ = π/4
(vii) x = a (θ − sin θ), y = a(1 − cos θ) at θ = π/2
(viii) y = (sin 2x + cot x + 2)2 at x = π/2
(ix) x2 + 3y + y2 = 5 at (1, 1)
(x) xy = 6 at (1, 6)

Answer:

i y=x3=x32dydx=32x12=32xWhen x=4,  y=x3=64=8Now,Slope of the tangent=dydx4, 8=324=3Slope of the normal=-1dydx4, 8=-13

ii y=x=x12dydx=12x-12=12xWhen x=9,  y=x=9=3Now,Slope of the tangent=dydx9, 3=129=16Slope of the normal=-1dydx9, 3=-116=-6

iii y=x3-xdydx=3x2-1When x=2,  y=x3-x=23-2=6Now,Slope of the tangent=dydx2, 6=322-1=11Slope of the normal=-1dydx2, 6=-111

iv y=2x2+3 sin xdydx=4x+3 cos xWhen x=0, y=2x2+3 sin x=202+3 sin 0=0Now,Slope of the tangent=dydx0, 0=40+ 3 cos 0=3Slope of the normal=-1dydx0, 0=-13


v x=aθ-sin θdxdθ=a1-cos θ y=a1+cos θ dydθ=a-sin θ dydx=dydθdxdθ=a-sin θa1-cos θ=-2 sin θ2 cos θ22 sin2θ2=-cot θ2Now,Slope of the tangent=dydxθ=-π2=-cot -π22=-cot -π4=1Slope of the normal=-1dydxθ=-π2=-11=-1

vi x=a cos3 θdxdθ=-3a cos2 θ sin θy=a sin3 θ dydθ=3a sin2 θ cos θ dydx=dydθdxdθ=3a sin2 θ cos θ-3a cos2 θ sin θ=-tan θNow,Slope of the tangent=dydxθ=π4=-tan π4=-1Slope of the normal=-1dydxθ=π4=-1-1=1


vii x=aθ-sin θdxdθ=a1-cos θ y=a1-cos θ dydθ=asin θ dydx=dydθdxdθ=asin θa1-cos θ=2 sin θ2 cos θ22 sin2θ2=cot θ2Now,Slope of the tangent=dydxθ=π2=cot π22=cot π4=1Slope of the normal=-1dydxθ=π2=-11=-1

viii y=sin 2x+cot x+22dydx=2 sin 2x+cot x+2 2cos 2x-cosec2xNow,Slope of the tangent=dydxx=π2=2 sin 2π2+cot π2+2 2cos 2π2-cosec2 π2=2 0+0+2 -2-1=-12Slope of the normal=-1dydxx=π2=-1-12=112

ix x2+3y+y2=5On differentiating both sides w.r.t. x, we get2x+3dydx+2y dydx=0dydx3+2y=-2xdydx=-2x3+2yNow,Slope of the tangent=dydx1, 1=-2x3+2y=-23+2=-25Slope of the normal=-1dydx1, 1=-1-25=52

x xy=6On differentiating both sides w.r.t. x, we getxdydx+y=0xdydx=-ydydx=-yxNow,Slope of the tangent=dydx1, 6=-yx=-61=-6Slope of the normal=-1dydx1, 6=-1-6=16

Page No 16.10:

Question 2:

Find the values of a and b if the slope of the tangent to the curve xy + ax + by = 2 at (1, 1) is 2.

Answer:

Given:xy+ax+by=2 ... 1On differentiating both sides w.r.t. x, we getxdydx+y+a+bdydx=0dydxx+b=-a-ydydx=-a-yx+bNow,dydx1, 1=2-a-11+b=2-a-1=2+2b-a=3+2ba=-3+2bOn substituting a=-3+2b,  x=1 and y=1 in eq. (1), we get1-3+2b+b=21-3-2b+b=2b=-4and a=-3+2b=-3-8=5∴ a=5 and b=-4

Page No 16.10:

Question 3:

If the tangent to the curve y = x3 + ax + b at (1, − 6) is parallel to the line xy + 5 = 0, find a and b.

Answer:

Given: x-y+5=0y=x+5dydx=1Now, y=x3+ax+b    ...1dydx=3x2+aSlope of the tangent at 1, -6 = Slope of the given linedydx1, -6=13+a=1a=-2On substituting a=-2,  x=1 and y=-6 in eq. (1), we get-6=1-2+bb=-5 a=-2 and b=-5

Page No 16.10:

Question 4:

Find a point on the curve y = x3 − 3x where the tangent is parallel to the chord joining (1, −2) and (2, 2).

Answer:

Let (x1, y1) be the required point.

Slope of the chord=y2-y1x2-x1=2+22-1=4y=x3-3xdydx=3x2-3  ...1Slope of the tangent=dydxx1, y1=3x12-3It is given that the tangent and the chord are parallel.∴ Slope of the tangent = Slope of the chord3x12-3=43x12=7x12=73x1=±73=73 or -73Case 1When x1=73On substituting this in eq. (1), we get y1=73 3-373 =7373 -373 =-2373  x1, y1=73, -2373 Case 2When x1=-73On substituting this in eq. (1), we get y1=-73 3-3-73 =-7373 +373 =2373  x1, y1=-73, 2373 

Page No 16.10:

Question 5:

Find the points on the curve y = x3 − 2x2 − 2x at which the tangent lines are parallel to the line y = 2x − 3.

Answer:

Let (x1, y1) be the required point.
Given:

y=2x-3 Slope of the line= dydx=2y=x3-2x2-2xSince x1y1 lies on curve, y1=x13-2x12-2x1    ...1dydxx1,y1=3x12-4x1-2It is given that the tangent and the given line are parallel.∴ Slope of the tangent = Slope of the given line3x12-4x1-2=23x12-4x1-4=03x12-6x1+2x1-4=03x1 x1-2 +2 x1-2=0x1-2 3x1+2=0x1=2 or x1=-23Case 1When x1=2On substituting the value of x1 in eq. (1), we get y1=8-8-4=-4 x1, y1=2, -4 Case 2When x1=-23On substituting the value of x1 in eq. (1), we get y1=-827-89+43=-8-24+3627=427 x1, y1=-23, 427

Page No 16.10:

Question 6:

Find the points on the curve y2 = 2x3 at which the slope of the tangent is 3.

Answer:

Let (x1, y1) be the required point.
Given:
y2=2x3 Since x1y1 lies on a curve, y12=2x13         ....12ydydx=6x2dydx=6x22y=3x2ySlope of the tangent at x, y=3x12y1Slope of the tangent=3       [Given] 3x12y1=3               ....2y1=x12On substituting the value of y1 in eq. (1), we getx14=2x13x13 x1-2=0x1=0, 2Case 1  When x1=0, y1=x2=0. Thus, we get the point 0, 0. But, it does not satisfy eq. (2). So, we can ignore (0, 0).Case 2 When x1=2, y1=x12=4. Thus, we get the point 2, 4.

Page No 16.10:

Question 7:

Find the points on the curve xy + 4 = 0 at which the tangents are inclined at an angle of 45° with the x-axis.

Answer:

Let the required point be (x1, y1).
Slope of the tangent at this point = tan 45° = 1
Given:
xy+4=0 ... 1Since the point satisfies the above equation,x1y1+4=0       ...2On differentiating equation 2 both sides with respect to x, we get xdydx+y=0dydx=-yxSlope of the tangent at x1, y1 = dydxx, y=-y1x1Slope of the tangent =1       [Given] -y1x1=1x1=-y1On substituting the value of x1 in eq. (2), we get-y12+4=0y12=4y1=±2Case 1When  y1=2, x1=-y1=-2∴ (x1, y1) = (-2, 2)Case 2When y1=-2, x1=-y1=2∴  x1, y1 = (2, -2)

Page No 16.10:

Question 8:

Find the point on the curve y = x2 where the slope of the tangent is equal to the x-coordinate of the point.

Answer:

Let the required point be (x1, y1).
Given:
y=x2Point x1, y1 lies on a curve. y1=x12           ...1Now,y=x2 dydx=2xSlope of the tangent at x1, y1 = dydxx1, y1=2x1Slope of the tangent =x coordinate of the point       [Given] 2x1=x1This happens only when x1= 0.On putting x1=0 in eq. 1, we gety1=x12=02=0Thus, the required point is 0, 0.

Page No 16.10:

Question 9:

At what points on the circle x2 + y2 − 2x − 4y + 1 = 0, the tangent is parallel to x-axis?

Answer:

Let the required point be (x1, y1).
We know that the slope of the x-axis is 0.
Given:

x2+y2-2x-4y+1=0 x1,y1 lies on a curve. x12+y12-2x1-4y1+1=0       ...1Now, x2+y2-2x-4y+1=0 2x+2y dydx-2-4dydx=0dydx 2y-4=2-2xdydx=2-2x2y-4=1-xy-2Slope of the tangent at x1, y1=dydxx1, y11-x1y1-2       ...(2)Slope of the tangent = 0      [Given]1-x1y1-2=01-x1=0x1=1On substituting the value of x1 in eq. (1), we get1+y12-2-4y1+1=0y12-4y1=0y1y1-4=0y1=0, 4Thus, the required points are (1, 0) and (1, 4).

Page No 16.10:

Question 10:

At what point of the curve y = x2 does the tangent make an angle of 45° with the x-axis?

Answer:

Let the required point be (x1, y1).
The tangent makes an angle of 45o with the x-axis.
∴ Slope of the tangent = tan 45o = 1

Since, the point lies on the curve.Hence, y12=x1Now, y2=x2ydydx=1dydx=12ySlope of the tangent = dydxx1, y1=12y1Given:12y1=12y1=1y1=12Now,x1=y12=122=14x1, y1=14, 12

Page No 16.10:

Question 11:

Find the points on the curve y = 3x2 − 9x + 8 at which the tangents are equally inclined with the axes.

Answer:

Let (x1, y1) be the required point.
It is given that the tangent at this point is equally inclined to the axes. It means that the angle made by the tangent with the x-axis is ±45°.
∴ Slope of the tangent = tan (±45) = ± 1  ...(1)

Since, the point lies on the curve.Hence, y1=3x12-9x1+8 Now, y=3x2-9x+8dydx=6x-9Slope of the tangent at x1, y1=dydxx1, y1=6x1-9     ...(2)From eq. (1) and eq. (2), we get6x1-9=±16x1-9=1 or 6x1-9=-16x1=10 or 6x1=8x1=106=53 or x1=86=43Also,y1=3532-953+8 or y1=3432-943+8y1=253-453+8 or y1=163-363+8y1=43 or y1=43Thus, the required points are 53, 43 and 43, 43.

Page No 16.10:

Question 12:

At what points on the curve y = 2x2x + 1 is the tangent parallel to the line y = 3x + 4?

Answer:

Let (x1, y1) be the required point.
The slope of line y = 3x + 4 is 3.

Since, the point lies on the curve.Hence, y1=2x12-x1+1Now, y=2x2-x+1dydx=4x-1Now,Slope of the tangent at x1, y1=dydxx1, y1=4x1-1Slope of the tangent at x1, y1= Slope of the given line       [Given] 4x1-1=34x1=4x1=1andy1=2x12-x1+1=2-1+1=2Thus, the required point is 1, 2.

Page No 16.10:

Question 13:

Find the point on the curve y = 3x2 + 4 at which the tangent is perpendicular to the line whose slop is -16.

Answer:

Let (x1, y1) be the required point.
Slope of the given line = -16
∴ Slope of the line perpendicular to it = 6

Since, the point lies on the curve.Hence, y1=3x12+4Now,  y=3x2+4 dydx=6xNow,Slope of the tangent at x1, y1=dydxx1, y1=6x1Slope of the tangent at x1, y1= Slope of the given line          [Given] 6x1=6x1=1andy1=3x12+4=3+4=7Thus, the required point is 1, 7.



Page No 16.11:

Question 14:

Find the points on the curve x2 + y2 = 13, the tangent at each one of which is parallel to the line 2x + 3y = 7.

Answer:

Let (x1, y1) represent the required point.
The slope of line 2x + 3y = 7 is -23.

Since, the point lies on the curve.Hence, x12+y12=13      ...1Now, x2+y2=13On differentiating both sides w.r.t. x, we get2x+2ydydx=0dydx=-xySlope of the tangent at x1, y1=dydxx1, y1=-x1y1Slope of the tangent at x1, y1= Slope of the given line    [Given]-x1y1=-23x1=2y13      ...2From eq. (1), we get2y132+y12=1313y129=13y12=9y1=±3y1=3 or y1=-3andx1=2 or x1=-2         [From eq. (2)]Thus, the required points are 2, 3 and -2, -3.

Page No 16.11:

Question 15:

Find the points on the curve 2a2y = x3 − 3ax2 where the tangent is parallel to x-axis.

Answer:

Let (x1, y1) represent the required points.
The slope of the x-axis is 0.
Here,
2a2y=x3-3ax2 Since, the point lies on the curve.Hence, 2a2y1=x13-3ax12     ...1Now, 2a2y=x3-3ax2 On differentiating both sides w.r.t. x, we get2a2dydx=3x2-6axdydx=3x2-6ax2a2Slope of the tangent at x1, y1=dydxx1, y1=3x12-6ax12a2Given:Slope of the tangent at x1, y1= Slope of the x-axis3x12-6ax12a2=03x12-6ax1=0x1 3x1-6a=0x1=0 or x1= 2aAlso,2a2y1=0 or 2a2y1=8a3-12a3      [From eq. (1)]y1=0 or y1=-2aThus, the required points are 0, 0 and 2a, -2a.

Page No 16.11:

Question 16:

At what points on the curve y = x2 − 4x + 5 is the tangent perpendicular to the line 2y + x = 7?

Answer:

Let (x1, y1) be the required point.
Slope of the given line = -12
Slope of the line perpendicular to this line = 2

Since, the point lies on the curve.Hence, y1=x12-4x1+5      ...1Now, y=x2-4x+5  dydx=2x-4Now,Slope of the tangent at x1, y1=dydxx1, y1=2x1-4Slope of the tangent at x1, y1=Slope of the given line        [Given]2x1-4=22x1=6x1=3Also,y1=9-12+5=2     [From eq. (1)]Thus, the required point is 3, 2.

Page No 16.11:

Question 17:

Find the points on the curve x24+y225=1 at which the tangents are parallel to the (i) x-axis (ii) y-axis.

Answer:

(i) The slope of the x-axis is 0.
Now, let (x1, y1) be the required point.
Since, the point lies on the curve.Hence, x124+y1225=1     ...1       Now, x24+y225=1      2x4+2y25dydx=02y25dydx=-x2dydx=-25x4yNow,Slope of the tangent at x1, y1=dydxx1, y1=-25x14y1Slope of the tangent at x1, y1=Slope of the x-axis     [Given] -25x14y1=0x1=0Also,0+y1225=1       [From eq. (1)]y12=25y1=±5Thus, the required points are 0, 5 and 0, -5.

(ii) The slope of the y-axis is .
Now, let (x1, y1) be the required point.
Since, the point lies on the curve.Hence, x124+y1225=1     ...1Now, x24+y225=1   2x4+2y25dydx=02y25dydx=-x2dydx=-25x4yNow,Slope of the tangent at x1, y1=dydxx1, y1=-25x14y1Slope of the tangent at x1, y1=Slope of the y-axis     [Given] -25x14y1=4y1-25x1=0y1=0Also,x124=1          [From eq. (1)]x12=4x1=±2Thus, the required points are 2, 0 and -2, 0.

Page No 16.11:

Question 18:

Find the points on the curve x2 + y2 − 2x − 3 = 0 at which the tangents are parallel to the x-axis.

Answer:

Let (x1, y1) be the required point.

Since the point lie on the curve.Hence x12+y12-2x1-3=0   ...1Now, x2+y2-2x-3=0 2x+2y dydx-2=0dydx=2-2x2y=1-xyNow,Slope of the tangent = dydxx1, y11-x1y1Slope of the tangent = 0    (Given)1-x1y1=01-x1=0x1=1From (1), we getx12+y12-2x1-3=01+y12-2-3=0y12-4=0y1=±2Hence, the points are 1, 2 and 1, -2.

Page No 16.11:

Question 19:

Find the points on the curve x29+y216=1 at which the tangents are (i) parallel to x-axis (ii) parallel to y-axis.

Answer:

(i) The slope of the x-axis is 0.
Now, let (x1, y1) be the required point.
Since, the point lies on the curve.Hence, x129+y1216=1      ...1Now, x29+y216=1   2x9+2y16dydx=0y16dydx=-x9dydx=-16x9yNow,Slope of the tangent at x, y=dydxx1, y1=-16x19y1Slope of the tangent at x, y= Slope of the x-axis         [Given] -16x19y1=0x1=00+y1216=1        [From eq. (1)]Also,y12=16y1=±4Thus, the required points are 0, 4 and 0, -4.

(ii) The slope of the y-axis is .
Let (x1, y1) be the required point.
Given:
Since, the point lies on the curve.Hence, x129+y1216=1      ...1x29+y216=1   2x9+2y16dydx=0y16dydx=-x9dydx=-16x9yNow,Slope of the tangent at x, y=dydxx1, y1=-16x19y1Slope of the tangent at x1, y1 = Slope of the y-axis       [Given]-16x19y1=9y1-16x1=0y1=0x129+0=1      [From eq. (1)]x12=9x1=±3Thus, the required points are 3, 0 and -3, 0.

Page No 16.11:

Question 20:

Who that the tangents to the curve y = 7x3 + 11 at the points x = 2 and x = −2 are parallel.

Answer:

Given: y=7x3+11 dydx=21x2Now,Slope of the tangent at (x=2) = dydxx=2=21 22=84Slope of the tangent at (x=-2) dydxx=-2=21 -22=84

Both slopes are the same. Hence, the tangents at points x = 2 and x = −2 are parallel.

Page No 16.11:

Question 21:

Find the points on the curve y = x3 where the slope of the tangent is equal to the x-coordinate of the point.

Answer:

Let (x1, y1) be the required point.
x coordinate of the point is x1.

Since, the point lies on the curve.Hence, y1=x13          ... 1Now, y=x3dydx=3x2Slope of tangent at x, y=dydxx1, y1=3x12Given thatSlope of tangent at x1, y1= x co-ordinate of the point3x12=x1x1 3x1-1=0x1=0 or x1=13y1=03 or y1=133 (From (1))y1=0 or y1=127So, the points are x1, y10, 013, 127



Page No 16.27:

Question 1:

Find the equation of the tangent to the curve x+y=a, at the point a24, a24.

Answer:

x+y=aDifferentiating both sides w.r.t. x,12x+12ydydx=0dydx=-yxGiven x1, y1=a24, a24Slope of tangent, m = dydxa24, a24=-a24a24=-1Equation of tangent is,y-y1=m x-x1y-a24=-1x-a24y-a24=-x+a24x+y=a22

Page No 16.27:

Question 2:

Find the equation of the normal to y = 2x3x2 + 3 at (1, 4).

Answer:

y=2x3-x2+3Differentiating both sides w.r.t. x,dydx=6x2-2xSlope of tangent = dydx1, 4=6 12-2 1=4Slope of normal = -1Slope of tangent=-14Given x1, y1=1, 4Equation of normal is,y-y1=m x-x1y-4=-14 x-14y-16=-x+1x+4y=17

Page No 16.27:

Question 3:

Find the equations of the tangent and the normal to the following curves at the indicated points.

(i) x4 − bx3 + 13x2 − 10x + 5 at (0, 5)                  [NCERT]
(ii) x4 − 6x3 + 13x2 − 10x + 5 at x = 1                  [NCERT, CBSE 2011]
(iii) x2 at (0, 0)                  [NCERT]
(iv) y = 2x2 − 3x − 1 at (1, −2)
(v) y2=x34-xat 2, -2
(vi) y = x2 + 4x + 1 at x = 3                       [CBSE 2004]
(vii) x2a2+y2b2=1 at acosθ, bsinθ
(viii) x2a2-y2b2=1 at asecθ, btanθ
(ix) y2 = 4ax at am2,2am
(x) c2 x2+y2=x2 y2  at ccosθ, csinθ
(ix) xy = c2 at ct,ct
(xii) x2a2+y2b2=1 at x1, y1
(xiii) x2a2-y2b2=1 at x0, y0             [NCERT]
(xiv) x23+y23 = 2 at (1, 1)              [NCERT]
(xv) x2 = 4y at (2, 1)
(xvi) y2 = 4x at (1, 2)                      [NCERT]
(xvii) 4x2 + 9y2 = 36 at (3cosθ, 2sinθ)              [CBSE 2011]
(xviii) y2 = 4ax at (x1, y1)               [CBSE 2012]
(xix) x2a2-y2b2=1 at 2a,b          [CBSE 2014]

Answer:

(i)
y=x4-bx3+13x2-10x+5Differentiating both sides w.r.t. x,dydx=4x3-3bx2+26x-10Slope of tangent, m = dydx0, 5=-10Given x1, y1=0, 5Equation of tangent is,y-y1=m x-x1y-5=-10x-0y-5=-10xy+10x-5=0

Equation of normal is,y-y1=-1m x-x1y-5=110 x-010y-50=xx-10y+50=0

(ii)
y=x4-6x3+13x2-10x+5When x=1 , y=1-6+13-10+5=3So, x1, y1=1, 3Now, y=x4-6x3+13x2-10x+5Differentiating both sides w.r.t. x,dydx=4x3-18x2+26x-10Slope of tangent, m = dydx1,3=4-18+26-10=2Equation of tangent is,y-y1=2 x-x1y-3=2x-1y-3=2x-22x-y+1=0

Equation of normal is,y-y1=-1m x-x1y-3=-12 x-13y-6=-x+1x+3y-7=0

(iii)
y=x2Differentiating both sides w.r.t. x,dydx=2xGiven x1, y1=0, 0Slope of tangent, m = dydx0, 0=2 0=0Equation of tangent is,y-y1=mx-x1y-0=0 x-0y=0

Equation of normal is,y-y1=-1m x-x1y-0=-10 x-0x=0

(iv)
y=2x2-3x-1Differentiating both sides w.r.t. x,dydx=4x-3Given x1, y1=1, -2Slope of tangent, m = dydx1, -2=4-3=1Equation of tangent is,y-y1=mx-x1y+2=1 x-1y+2=x-1x-y-3=0

Equation of normal is,y-y1=-1m x-x1y+2=-1 x-1y+2=-x+1x+y+1=0

(v)
y2=x34-xDifferentiating both sides w.r.t. x,2y dydx=4-x3x2-x3-14-x2=12x2-3x3+x34-x2=12x2-2x34-x2dydx=12x2-2x32y 4-x2Given x1, y1=2, -2Slope of tangent, m = dydx2, -2=48-16-16=-2Equation of tangent is,y-y1=mx-x1y+2=-2 x-2y+2=-2x+42x+y-2=0

Equation of normal is,y-y1=-1m x-x1y+2=12 x-22y+4=x-2x-2y-6=0

(vi)
y=x2+4x+1Differentiating both sides w.r.t. x,dydx=2x+4When x =3, y=9+12+1=22 So, x1, y1=3, 22Slope of tangent, m = dydxx=3=10Equation of tangent is,y-y1=m x-x1y-22=10x-3y-22=10x-3010x-y-8=0

Equation of normal is,y-y1=-1m x-x1y-22=-110 x-310y-220=-x+3x+10y-223=0

(vii)
x2a2+y2b2=1Differentiating both sides w.r.t. x,2xa2+2yb2dydx=02yb2dydx=-2xa2dydx=-xb2ya2Slope of tangent, m = dydxa cos θ, b sin θ=-a cos θ b2b sin θ a2=-b cos  θa sin  θGiven x1, y1=a cos θ, b sin θEquation of tangent is,y-y1=m x-x1y-b sin θ=-b cos  θa sin  θx-a cos θay sin θ-ab sin2 θ=-bx cos θ+ab cos2 θbx cos θ+ay sin θ=abDividing by ab,xacos θ +ybsin θ=1

Equation of normal is,y-y1=-1m x-x1y-b sin θ=a sin θb cos θx-a cos θby cos θ-b2sin θ cos θ=ax sin θ-a2sin θ cos θax sin θ-by cos θ=a2-b2sin θ cos θDividing by sin θ cos θ,ax sec θ-by cosec θ=a2-b2

(viii)
x2a2-y2b2=1Differentiating both sides w.r.t. x,2xa2-2yb2dydx=02yb2dydx=2xa2dydx=xb2ya2Slope of tangent, m = dydxa sec θ, b tan θ=a sec θ b2b tan θ a2=b a sin  θGiven x1, y1=a sec θ, b tan θEquation of tangent is,y-y1=m x-x1y-b tan θ=b a sin  θx-a sec θay sin θ -ab sin2θcos θ=bx-abcos θay sin θ cos θ-ab sin2θcos θ=bx cos θ -abcos θay sin θ cos θ-ab sin2θ=bx cos θ -abbx cos θ-ay sin θ cos θ=ab 1-sin2 θ bx cos θ-ay sin θ cos θ=ab cos2θDividing by ab  cos2 θ,xasec θ -ybtan θ=1

Equation of normal is,y-y1=-1m x-x1y-b tan θ=-a sin θbx-a sec θyb-b2tan θ=-ax sin θ+a2 tanθax sin θ+by=a2+b2tan θDividing by tan θ,ax cos θ+by cot θ=a2+b2

(ix)
y2=4axDifferentiating both sides w.r.t. x,2y dydx=4adydx=2ayGiven x1, y1=am2, 2amSlope of tangent = dydxam2, 2am=2a2am=mEquation of tangent is,y-y1=m x-x1y-2am=m x-am2my-2am=mm2x-am2my-2a=m2x-am2x-my+a=0

Equation of normal is,y-y1=1Slope of tangent x-x1y-2am=-1mx-am2my-2am=-1mm2x-am2m3y-2am2=-m2x+am2x+m3y-2am2-a=0

(x)
c2x2+y2=x2y2Differentiating both sides w.r.t. x,2x c2+2y c2dydx=x2 2y dydx+2xy2dydx2y c2-2x2y=2xy2-2xc2dydx=xy2-xc2yc2-x2ySlope of tangent, m = dydxccos θ, csin θ=c3cos θ sin2θ  -c3cos θc3sinθ-c3cos2θ sinθ=1-sin2θcosθ sin2θcos2θ-1cos2θ sinθ=cos2θcos θ sin2θ×cos2θ sinθ-sin2θ=-cos3θsin3θGiven x1, y1=ccos θ, csin θEquation of tangent is,y-y1=m x-x1y-csin θ=-cos3θsin3θ x-ccos θysinθ-csinθ=-cos3θsin3θx cosθ-ccosθsin2θy sinθ -c=-cos2θxcosθ-cysin3θ-csin2θ=-xcos3θ+ccos2θxcos3θ+ysin3θ=csin2θ+cos2θxcos3θ+ysin3θ=c

Equation of normal is,y-y1=-1m x-x1y-csin θ=sin3θcos3θx-ccos θcos3θy-csin θ=sin3θx-ccos θycos3θ-c cos3θsinθ=xsin3θ-c sin3θcosθxsin3θ-ycos3θ=c sin3θcosθ-c cos3θsinθxsin3θ-ycos3θ=csin4θ-cos4θcosθ sinθxsin3θ-ycos3θ=csin2θ+cos2θsin2θ-cos2θcosθ sinθsin3θ-ycos3θ=2c-cos2θ-sin2θ2cosθ sinθsin3θ-ycos3θ=2c-cos 2θsin2θsin3θ-ycos3θ=-2c cot2θsin3θ-ycos3θ+2c cot2θ=0

(xi)
xy=c2Differentiating both sides w.r.t. x,xdydx+y=0dydx=-yxGiven x1, y1=ct, ctSlope of tangent, m = dydxct, ct=-ctct=-1t2Equation of tangent is,y-y1=m x-x1y-ct=-1t2 x-ctyt-ct=-1t2 x-ctyt2-ct=-x+ctx+yt2=2ct

Equation of normal is,y-y1=-1m x-x1y-ct=t2x-ctyt-c=t3x-ct4xt3-yt=ct4-c

(xii)
x2a2+y2b2=1Differentiating both sides w.r.t. x,2xa2+2yb2dydx=02yb2dydx=-2xa2dydx=-xb2ya2Slope of tangent, m = dydxx1, y1=-x1b2y1a2Equation of tangent is,y-y1=m x-x1y-y1=-x1b2y1a2x-x1yy1a2-y12a2=-xx1b2+x12b2xx1b2+yy1a2=x12b2+y12a2 ... 1Since x1, y1 lies on the given curve.Therefore,x12a2+y12b2=1x12b2+y12a2a2b2=1x12b2+y12a2=a2b2Substituting this in (1), we getxx1b2+yy1a2=a2b2Dividing this by a2b2,xx1a2+yy1b2=1

Equation of normal is,y-y1=m x-x1y-y1=y1a2x1b2x-x1yx1b2-x1y1b2=xy1a2-x1y1a2xy1a2-yx1b2=x1y1a2-x1y1b2xy1a2-yx1b2=x1y1a2-b2Dividing by x1y1a2xx1-b2yy1=a2-b2

(xiii)
x2a2-y2b2=1Differentiating both sides w.r.t. x,2xa2-2yb2dydx=02yb2dydx=2xa2dydx=xb2ya2Slope of tangent, m = dydxx0, y0=x0b2y0a2Equation of tangent is,y-y1=m x-x1y-y0=x0b2y0a2x-x0yy0a2-y02a2=xx0b2-x02b2xx0b2-yy0a2=x02b2-y02a2 ... 1Since x0, y0 lies on the given curve,x02a2-y02b2=1x02b2-y02a2=a2b2Substituting this in (1), we getxx0b2-yy0a2=a2b2Dividing this by a2b2xx0a2-yy0b2=1

Equation of normal is,y-y1=m x-x1y-y0=-y0a2x0b2x-x0yx0b2-x0y0b2=-xy0a2+x0y0a2xy0a2+yx0b2=x0y0a2+x0y0b2xy0a2+yx0b2=x0y0a2+b2Dividing by x0y0a2xx0+b2yy0=a2+b2

(xiv)
x23+y23=2Differentiating both sides w.r.t. x,23x-13+23y-13dydx=0dydx=-x-13y-13=-y13x13Slope of tangent, m = dydx1, 1-11=-1Given x1, y1=1, 1Equation of tangent is,y-y1=m x-x1y-1=-1x-1y-1=-x+1x+y-2=0

Equation of normal is,y-y1=-1m x-x1y-1=1x-1y-1=x-1y-x=0

(xv)
x2=4yDifferentiating both sides w.r.t. x,2x=4dydxdydx=x2Slope of tangent, m = dydx2, 1=22=1Given x1, y1=2, 1Equation of tangent is,y-y1=m x-x1y-1=1x-2y-1=x-2x-y-1=0

Equation of normal is,y-y1=-1m x-x1y-1=-1x-2y-1=-x+2x+y-3=0

(xvi)
y2=4xDifferentiating both sides w.r.t. x,2y dydx=4dydx=2ySlope of tangent, m = dydx1, 2=22=1Given x1, y1=1, 2Equation of tangent is,y-y1=m x-x1y-2=1x-1y-2=x-1x-y+1=0

Equation of normal is,y-y1=-1m x-x1y-2=-1x-1y-2=-x+1x+y-3=0

(xvii) Equation of tangent:
4x2+9y2=36Differentiating both sides w.r.t. x,8x+18y dydx=018y dydx=-8xdydx=-8x18y=-4x9ySlope of tangent, m = dydx3 cosθ, 2 sinθ=-12cosθ18sinθ=-2 cosθ3 sinθGiven x1, y1=3 cosθ, 2 sinθEquation of tangent is,y-y1=m x-x1y-2 sinθ=-2 cosθ3 sinθx-3 cosθ3y sinθ-6sin2θ=-2x cosθ+6cos2θ2x cosθ+3y sinθ=6cos2θ+sin2θ2x cosθ+3y sinθ=6

Equation of normal is,y-y1=-1m x-x1y-2 sinθ=3 sinθ2 cosθx-3 cosθ2y cosθ-4 sinθ cosθ=3x sinθ-9 sinθ cosθ3x sinθ-2y cosθ-5sinθ cosθ=0

(xviii)
y2=4axDifferentiating both sides w.r.t. x,2ydydx=4adydx=2ayAt x1, y1Slope of tangent = dydxx1, y1=2ay1=mEquation of tangent is,y-y1=mx-x1y-y1=2ax-x1y1yy1-y12=2ax-2ax1yy1-4ax1=2ax-2ax1yy1=2ax+2ax1yy1=2ax+x1

Equation of normal is,y-y1=1Slope of tangent x-x1y-y1=-1mx-x1y-y1=-y12ax-x1

(xix)
x2a2-y2b2=1Differentiating both sides w.r.t. x,2xa2-2yb2dydx=02yb2dydx=2xa2dydx=xb2ya2Slope of tangent, m = dydx2a,b=2ab2ba2=2baEquation of tangent is,y-y1=mx-x1y-b=2bax-2aay-ab=2bx-2ab2bx-ay=ab2xa-yb=1

Equation of normal is,y-y1=-1mx-x1y-b=-a2bx-2a2by-2b2=-ax+2a2ax+2by=2b2+2a2ax2+by=a2+b2



Page No 16.28:

Question 4:

Find the equation of the tangent to the curve x = θ + sin θ, y = 1 + cos θ at θ = π/4.

Answer:

x=θ+sin θ and y=1+cos θdxdθ=1+cos θ  and dydθ=-sin θdydx=dydθdxdθ=-sin θ1+cos θSlope of tangent=dydxθ=π4=-sin π41+cos π4=-121+12=-12+1=-12+1×2-12-1=1-2x1, y1=π4+sinπ4, 1+cos π4=π4+12, 1+12Equation of tangent is,y-y1=mx-x1y-1+12=1-2x-π4+12y-1-12=1-2x-π4-12

Page No 16.28:

Question 5:

Find the equations of the tangent and the normal to the following curves at the indicated points.

(i) x = θ + sinθy = 1 + cosθ at θ = π2
(ii) x=2 at21+t2, y=2 at31+t2at t=12
(iii) x = at2y = 2at at t = 1
(iv) x = asecty = btant at t
(v) x = a(θ + sinθ), y = a(1 − cosθ) at θ
(vi) 
x = 3cosθ − cos3θy = 3sinθ − sin3θ                    [NCERT EXEMPLAR]

Answer:

(i)
x=θ+sinθ and y=1+cosθdxdθ=1+cosθ and  dydθ=-sinθdydx=dydθdxdθ=-sinθ1+cosθSlope of tangent, m = dydxθ=π2=-sinπ21+cosπ2=-11+0=-1Now,  x1, y1=π2+sinπ2, 1+cosπ2=π2+1, 1Equation of tangent is,y-y1=m x-x1y-1=-1x-π2-12y-2=-2x+π+22x+2y-π-4=0

Equation of normal is,y-y1=-1m x-x1y-1=1 x-π2-12y-2=2x-π-22x-2y=π

(ii) Equation of tangent:
x=2at21+t2 and y=2at31+t2dxdt=1+t24at-2at22t1+t22=4at1+t22 anddydt=1+t26at2-2at32t1+t22=6at2+2at41+t22dydx=dydtdxdt=6at2+2at41+t224at1+t22=6at2+2at44atSlope of tangent, m = dydxt=12=3a2+a82a=12a+a82a=13a8×12a=1316Now, x1, y1=2at21+t2,2at31+t2 =a21+14, a41+14=a254, a454=2a5, a5Equation of tangent is,y-y1=m x-x1y-a5=1316x-2a55y-a5=13165x-2a55y-a=13165x-2a80y-16a=65x-26a65x-80y-10a=013x-16y-2a=0

Equation of normal is,y-y1=-1m x-x1y-a5=-1613 x-2a55y-a5=-16135x-2a55y-a=-16135x-2a65y-13a=-80x+32a80x+65y-45a=016x+13y-9a=0

(iii)
x=at2 and y=2atdxdt=2at and dydt=2adydx=dydtdxdt=2a2at=1tSlope of tangent, m = dydxt=1=11=1Now, x1, y1=a, 2aEquation of tangent is,y-y1=m x-x1y-2a=1x-ay-2a=x-ax-y+a=0

Equation of normal:

Equation of normal is,y-y1=m x-x1y-2a=-1 x-ay-2a=-x+ax+y=3a

(iv)
x=a sec t and y=b tan tdxdt=a sec t tan t  and dydt=b sec2tdydx=dydtdxdt=b sec2ta sec t tan t=bacosec tSlope of tangent, m = dydxt=t=bacosec tNow, x1, y1=a sec t, b tan tEquation of tangent is,y-y1=m x-x1y- b tan t=bacosec tx-a sec ty-b sin tcos t=ba sin tx-acos ty cos t-b sin tcos t=ba sin tx cos t-acos ty cos t-b sin t=ba sin tx cos t-aay sin t cos t-ab sin2t=bx cos t-abbx cos t-ay sin t cos t-ab1-sin2t=0bx cos t-ay sin t cos t=ab cos2tDividing by cos2t,bx sec t-ay tan t=ab

Equation of normal is,y-y1=m x-x1y- b tan t=-absin tx-a sec ty- b sin tcos t=-absin tx-acos ty cos t-b sin tcos t=-absin tx cos t-acos ty cos t-b sin t=-absin tx cos t-aby cos t-b2sin t=-ax sin t cos t+a2sin tax sin t cos t+by cos t=a2+b2sin tDividing both sides by sin t,ax cos t+by cot t=a2+b2

(v)
x=aθ+sinθ and y=a1-cosθdxdθ=a1+cosθ and dydθ=asinθdydx=dydθdxdθ=asinθa1+cosθ=sinθ1+cosθ=2sinθ2cosθ22cos2θ2=tanθ2  ...1Slope of tangent, m = dydxθ=tanθ2Now, x1, y1=aθ+sinθ, a1-cosθ Equation of tangent is,y-y1=m x-x1y-a1-cosθ=tanθ2x-aθ+sinθy-a2 sin2θ2=xtanθ2-aθtanθ2-atanθ2sinθy-a2 sin2θ2=xtanθ2-aθtanθ2-a2sinθ2cosθ22cos2θ22sinθ2cosθ2 (From (1))y-2a sin2θ2=x-aθtanθ2-2a sin2θ2y=x-aθtanθ2

Equation of normal is,y-a1-cosθ=-cotθ2x-aθ+sinθtan θ2y-a2 sin2θ2=-x+aθ+asinθtan θ2y-a2 1-cos2θ2=-x+aθ+asinθtan θ2y-2a+a 2 sin θ2 cos θ2=-x+aθ+asinθtan θ2y-2a+asinθ=-x+aθ+asinθtan θ2y-2a=-x+aθtan θ2y-2a+x-aθ=0

(vi)
x=3cosθ-cos3θ and y=3sinθ-sin3θdxdθ=-3sinθ+3cos2θsinθ=3sinθcos2θ-1=3sinθ-sin2θ=-3sin3θ and dydθ=3cosθ-3sin2θcosθ=3cosθ1-sin2θ=3cosθcos2θ=3cos3θdydx=3cos3θ-3sin3θ=-tan3θ  ...1Slope of tangent, m = dydxθ=-tan3θNow, x1, y1=3cosθ-cos3θ, 3sinθ-sin3θ Equation of tangent is,y-y1=m x-x1y-3sinθ-sin3θ=-tan3θx-3cosθ-cos3θy-3sinθ-sin3θ=-sin3θcos3θx-3cosθ-cos3θycos3θ-3sinθcos3θ-sin3θcos3θ=-xsin3θ+3sin3θcosθ+sin3θcos3θxsin3θ+ycos3θ=3sin3θcosθ+3sinθcos3θ+2sin3θcos3θxsin3θ+ycos3θ=3sin3θcosθsin2θ+cos2θ+2sin3θcos3θxsin3θ+ycos3θ=3sin3θcosθ+2sin3θcos3θx+ycot3θ=3cosθ+2cos3θ

Equation of normal is,y-3cosθ-cos3θ=-1mx-3sinθ-sin3θy-3cosθ+cos3θ=-1tan3θx-3sinθ+sin3θy-3cosθ+cos3θ=-cos3θsin3θx-3sinθ+sin3θysin3θ-3sin3θcosθ+sin3θcos3θ=-xcos3θ+3cos3θsinθ-cos3θsin3θxcos3θ+ysin3θ=3sin3θcosθ+3cos3θsinθ-2cos3θsin3θxcos3θ+ysin3θ=3sin3θcosθsin2θ+cos2θ-2cos3θsin3θxcos3θ+ysin3θ=3sin3θcosθ-2cos3θsin3θxcot3θ+y=3cosθ-2cos3θ

Page No 16.28:

Question 6:

Find the equation of the normal to the curve x2 + 2y2 − 4x − 6y + 8 = 0 at the point whose abscissa is 2.

Answer:

Abscissa means the horizontal co-ordiante of a point.
Given that abscissa = 2.
i.e., x = 2

x2+2y2-4x-6y+8=0 ... 1Differentiating both sides w.r.t. x,2x+4ydydx-4-6dydx=0dydx4y-6=4-2xdydx=4-2x4y-6=2-x2y-3When x=2, from (1), we get4+2y2-8-6y+8=02y2-6y+4=0y2-3y+2=0y-1y-2=0y=1 or y=2Case-1:  y=1Slope of tangent = dydx2, 1=0-1=0x1, y1=2, 1Equation of normal is,y-y1=-1m x-x1y-1=-10 x-2x-2=0x=2Case-2:  y=2Slope of tangent = dydx2, 2=01=0x1, y1=2, 2Equation of normal is,y-y1=-1m x-x1y-2=-10 x-2x-2=0x=2

In both cases, the equation of normal is x = 2

Page No 16.28:

Question 7:

Find the equation of the normal to the curve ay2 = x3 at the point (am2, am3).

Answer:

ay2=x3Differentiating both sides w.r.t. x,2ay dydx=3x2dydx=3x22aySlope of tangent = dydxam2, am3=3a2m42a2m3=3m2Given x1, y1=am2, am3Equation of normal is,y-y1=-1m x-x1y-am3=-23m x-am23my-3am4=-2x+2am22x+3my-am22+3m2=0

Page No 16.28:

Question 8:

The equation of the tangent at (2, 3) on the curve y2 = ax3 + b is y = 4x − 5. Find the values of a and b.

Answer:

The slope of the given line y = 4x − 5 is 4
y2=ax3+b ... 12y dydx=3ax2dydx=3ax22ySlope of tangent=dydx2, 3=12a6=2aGiven thatSlope of tangent= slope of given line2a=4a=2Substituting this and x = 2, y = 3 in (1), we get9=16+bb=-7Hencea=2 and b=-7

Page No 16.28:

Question 9:

Find the equation of the tangent line to the curve y = x2 + 4x − 16 which is parallel to the line 3xy + 1 = 0.

Answer:

Let (x0, y0) be the point of intersection of both the curve and the tangent.
y=x2+4x-16Since, x0,y0 lies on curve. Thereforey0=x02+4x0-16 ... 1Now, y=x2+4x-16dydx=2x+4Slope of tangent = dydxx0, y0=2x0+4Given that The tangent is parallel to the line So,Slope of tangent=slope of the given line2x0+4=32x0=-1x0=-12From (1),y0=14-2-16=-714Now, slope of tangent, =3x0, y0=-12, -714Equation of tangent isy-y0=m x-x0y+714=3x+124y+714=32x+124y+71=12x+612x-4y-65=0

Page No 16.28:

Question 10:

Find an equation of normal line to the curve y = x3 + 2x + 6 which is parallel to the line x + 14y + 4 = 0.

Answer:

Let (x1, y1) be a point on the curve where we need to find the normal.
Slope of the given line = -114
Since, the point lies on the curve.Hence, y1=x13+2x1+6 Now, y=x3+2x+6dydx=3x2+2Slope of the tangent=dydxx1, y1=3x12+2Slope of the normal=-1slope of the tangent==-13x12+2Given that,slope of the normal=slope of the given line-13x12+2=-1143x12+2=143x12=12x12=4x1=±2Case-1:  x1=2y1=x13+2x1+6=8+4+6=18x1, y1=2, 18Slope of the normal, m=-114Equation of normal is,y-y1=m x-x1y-18=-114x-214y-252=-x+2x+14y-254=0Case-2: x1=-2y1=x13+2x1+6=-8-4+6=-6x1, y1=-2, -6Slope of the normal, m=-114Equation of normal is,y-y1=m x-x1y+6=-114x+214y+84=-x-2x+14y+86=0

Page No 16.28:

Question 11:

Determine the equation(s) of tangent (s) line to the curve y = 4x3 − 3x + 5 which are perpendicular to the line 9y + x + 3 = 0.

Answer:

Let (x1, y1) be a point on the curve where we need to find the tangent(s).
Slope of the given line = -19

Since, tangent is perpendicular to the given line,
Slope of the tangent = -1-19=9
Let x1, y1 be the point where the tangent is drawn to this curve.Since, the point lies on the curve.Hence, y1=4x13-3x1+5 Now, y=4x3-3x+5dydx=12x2-3Slope of the tangent=dydxx1, y1=12x12-3Given that,slope of the tangent=slope of the perpendicular line12x12-3=912x12=12x12=1x1=±1Case-1: x1=1y1=4x13-3x1+5=4-3+5=6 x1, y1=1, 6Slope of the tangent=9Equation of tangent is,y-y1=m x-x1y-6=9x-1y-6=9x-99x-y-3=0Case-2: x1=-1y1=4x13-3x1+5=-4+3+5=4x1, y1=-1, 4Slope of the tangent=9Equation of tangent is,y-y1=m x-x1y-4=9x+1y-4=9x+99x-y+13=0

Page No 16.28:

Question 12:

Find the equation of a normal to the curve y = x loge x which is parallel to the line 2x − 2y + 3 = 0.

Answer:

Slope of the given line is 1
Let x1, y1 be the point where the tangent is drawn to the curve.Since, the point lies on the curve.Hence, y1= x1 loge x1            ... 1Now, y= x loge x dydx=x×1x+loge x 1=1+loge xSlope of tangent=1+loge x1Slope of normal = -1Slope of tangent=-11+loge x1Given thatSlope of normal = slope of the given line-11+loge x1=1-1=1+loge x1-2=loge x1x1=e-2=1e2Now, y1= e-2-2=-2e2         From (1)x1, y1=1e2, -2e2Equation of normal is,y+2e2=1 x-1e2y+2e2=x-1e2x-y=3e2x-y=3e-2

Page No 16.28:

Question 13:

Find the equation of the tangent line to the curve y = x2 − 2x + 7 which is (i) parallel to the line 2xy + 9 = 0 (ii) perpendicular to the line 5y − 15x = 13.

Answer:

(i) Slope of the given line is 2
Let x1, y1 be the point where the tangent is drawn to the curve y=x2-2x+7 Since, the point lies on the curve.Hence, y1=x12-2x1+7          ... 1Now, y=x2-2x+7dydx=2x-2Slope of tangent at point x1, y1=2x1-2Given thatSlope of tangent= Slope of the given line2x1-2=22x1=4x1=2Now, y1= 4-4+7=7x1, y1=2, 7Equation of tangent is,y-y1=m x-x1y-7=2 x-2y-7=2x-42x-y+3=0

(ii) Slope of the given line is 3
Slope of the line perpendicular to this line = -13

Let x1, y1 be the point where the tangent is drawn to the curve.Since, the point lies on the curve.Hence, y1=x12-2x1+7          ... 1Now, y=x2-2x+7dydx=2x-2Slope of tangent  at x1, y1=2x1-2Given thatSlope of tangent at x1, y1= Slope of the perpendicular line2x1-2=-136x1-6=-16x1=5x1=56Now, y1=2536-106+7=25-60+25236=21736x1, y1=56,21736Equation of tangent is,y-21736=-13 x-5636y-21736=-6x+51836y-217=-12x+1012x+36y-227=0

Page No 16.28:

Question 14:

Find the equations of all lines having slope 2 and that are tangent to the curve y=1x-3, x3.

Answer:

Slope of given tangent = 2
Let x1, y1 be the point where the tangent is drawn to this curve.Since, the point lies on the curve.Hence, y1=1x1-3Now, y=1x-3dydx=-1x-32Slope of tangent = dydx=-1x1-32Given that Slope of the tangent  = 2-1x1-32=2x1-32=-2x1-3=-2, which does not exist because 2 is negative.
So, there does not exist any such tangent.

Page No 16.28:

Question 15:

Find the equations of all lines of slope zero and that are tangent to the curve y=1x2-2x+3.

Answer:

Slope of the given tangent is 0.

Let x1,y1be a point where the tangent is drawn to the curve (1).Since, the point lies on the curve.Hence, y1=1x12-2x1+3... 1 Now, y=1x2-2x+3dydx=x2-2x+30-2x-21x2-2x+32=-2x+2x2-2x+32Slope of tangent=-2x1+2x12-2x1+32Given thatSlope of tangent = slope of the given line-2x1+2x12-2x1+32=0-2x1+2=02x1=2x1=1Now, y= 11-2+3=12       From1x1, y1=1, 12Equation of tangent is,y-y1=m x-x1y-12=0 x-1y=12

Page No 16.28:

Question 16:

Find the equation of the tangent to the curve y=3x-2 which is parallel to the 4x − 2y + 5 = 0.

Answer:

Slope of the given line is 2
Let x1,y1be the point where the tangent is drawn to the curve y=3x-2Since, the point lies on the curve.Hence, y1=3x1-2                 ... 1Now, y=3x-2dydx=323x-2Slope of tangent at  x1,y1 =323x1-2Given thatSlope of tangent = slope of the given line323x1-2=23=43x1-29=163x1-2916=3x1-23x1=916+2=9+3216=4116x1=4148Now, y1=12348-2=2748=916=34            From (1)x1, y1=4148, 34Equation of tangent is,y-y1=m x-x1y-34=2 x-41484y-34=248x-414824y-18=48x-4148x-24y-23=0

Page No 16.28:

Question 17:

Find the equation of the tangent to the curve x2 + 3y − 3 = 0, which is parallel to the line y = 4x − 5.

Answer:

Suppose (x1, y1) be the point of contact of tangent.
We can find the slope of the given line by differentiating the equation w.r.t  x
So, Slope of the line  = 4

Since,x1, y1 lies on the curve.Therefore,  x12+3y1-3=0      ... 1Now, x2+3y-3=02x+3dydx=0dydx=-2x3Slope of tangent,  m=dydxx1, y1=-2x13Given that tangent is parallel to the line, SoSlope of tangent, m = slope of the given line-2x13=4x1=-636+3y1-3=0(From (1))3y1=-33y1=-11x1, y1=-6, -11Equation of tangent is,y-y1=m x-x1y+11=4 x+6y+11=4x+244x-y+13=0



Page No 16.29:

Question 18:

Prove that xan+ybn=2 touches the straight line xa+yb=2 for all n ∈ N, at the point (a, b).

Answer:

Now, xan+ybn=2na xan-1+nbybn-1dydx=0nbybn-1dydx=-na xan-1dydx=-na xan-1×bnbyn-1=-babxayn-1Slope of tangent=dydxa, b=-bab*aa*bn-1=-ba ... (2)The equation of tangent isy-b=-bax-aya-ab=-xb+abxb+ya=2abxa+yb=2

So, the given line touches the given curve at the given point.

Page No 16.29:

Question 19:

Find the equation of the tangent to the curve x = sin 3t, y = cos 2t at t=π4.

Answer:

x=sin 3t and y=cos 2tdxdt=3 cos 3t and dydt=-2 sin 2tdydx=dydtdxdt=-2 sin 2t3 cos 3tSlope of tangent, m = dydxt=π4=--2 sin π23 cos 3π4=-2-32=223x1=sin 3×π4=12 and y1=cos 2×π4=0So, x1, y1=12, 0Equation of tangent is,y-y1=m x-x1y-0=223x-123y=22x-222x-3y-2=0

Page No 16.29:

Question 20:

At what points will be tangents to the curve y = 2x3 − 15x2 + 36x − 21 be parallel to x-axis? Also, find the equations of the tangents to the curve at these points.

Answer:

Slope of x - axis is 0
Let (x1, y1) be the required point.
y=2x3-15x2+36x-21Sincex1,y1  lies on the curve.Therefore y1=2x13-15x12+36x1-21    ... 1Now, y=2x3-15x2+36x-21dydx=6x2-30x+36Slope of tangent at x1, y1=dydxx1, y1= 6x12-30x1+36Given thatSlope of tangent at x, y= slope of the x-axis6x12-30x1+36=0x12-5x1+6=0x1-2x1-3=0x1=2 or x1=3Case-1: x1=2y1=16-60+72-21=7 (From (1))x1, y1=2, 7Equation of tangent is,y-y1=mx-x1y-7=0x-2y=7Case-2:  x1=3y1=54-135+108-21=6 (From (1))x1, y1=3, 6Equation of tangent is,y-y1=mx-x1y-6=0x-3y=6

Page No 16.29:

Question 21:

Find the equation of  the tangents to the curve 3x2y2 = 8, which passes through the point (4/3, 0).

Answer:

We have,

3x2 – y2 = 8                   ...(i)

Differentiating both sides w.r.t x, we get

6x-2ydydx=02ydydx=6xdydx=6x2ydydx=3xy

Let tangent at (h, k) pass through 43, 0.

Since, (h, k) lies on (i), we get

3h2-k2=8                   ...(ii)

Slope of tangent at (h, k) = 3hk

The equation of tangent at (h, k) is given by,

(y-k)=3hk(x-h)                    ...(iii)

Since, the tangent passess through 43, 0.

 (0-k)=3hk43-h-k=4hk-3h2k-k2=4h-3h2

8-3h2=4h-3h2          From ii8=4hh=2

Using (ii), we get

12-k2=8k2=4k=±2

So, the points on curve (i) at which tangents pass through 43, 0 are 2, ±2.

Now, from (iii), the equation of tangents are
(y-2)=62(x-2), or, 3x-y-4=0, and(y+2)=6-2(x-2), or, 3x+y-4=0



Page No 16.40:

Question 1:

Find the angle of intersection of the following curves:

(i) y2 = x and x2 = y             [NCERT EXEMPLAR]
(ii) y = x2 and x2 + y2 = 20
(iii) 2y2 = x3 and y2 = 32x
(iv) x2 y2 − 4x − 1 = 0 and x2 + y2 − 2y − 9 = 0
(v) x2a2+y2b2=1 and x2 + y2 = ab
(vi) x2 + 4y2 = 8 and x2 − 2y2 = 2
(vii) x2 = 27y and y2 = 8x
(viii) x2 + y2 = 2x and y2 = x
(ix) y = 4 − x2 and yx2                      [NCERT EXEMPLAR]

Answer:

​i  Given curves are,y2=x ... 1x2=y ... 2From these two equations, we getx22=xx4-x=0x x3-1=0x=0 or = 1Substituting the values of x in 2 we get,y=0 or y=1      x, y = 0, 0  or 1, 1Differenntiating (1) w.r.t. x,2y dydx=1dydx=12y            ...3Differenntiating (2) w.r.tx,2x = dydx                ...4Case -1:  x, y = 0, 0The tangent to curve is parallel to x-axis.Hence, the angle between the tangents to two curve at 0, 0 is a right angle.θ=π2Case -2: x, y = 1, 1From 3 we have, m1=12From 4 we have, m2=2 1=2Now,tan θ=m1-m21+m1m2=12-21+12×2=34θ=tan-1 34

ii Given curves are,y=x2                      ... 1x2+y2=20          ... 2From these two equations we get y+y2=20y2+y-20=0y+5y-4=0y=-5 or y=4Substituting the values of y in 1 we get,x2=-5 or x2=4   x=±2 and  x2 = -5  has no real solutionSo, x, y=2, 4 or -2, 4Differenntiating (1) w.r.t. x,dydx=2x               ...3Differenntiating (2) w.r.tx,2x +2y dydx=0dydx=-xy               ...4Case -1: x, y = 2, 4From 3 we have, m1=22=4From 4 we have,  m2=-24=-12Now, tan θ=m1-m21+m1m2=4+121+4 -12=92θ=tan-1 92Case -2:  x, y = -2, 4From 3 we have, m1=2-2=-4From 4 we have, m2=24=12Now, tan θ=m1-m21+m1m2=-4-121-4 12=92θ=tan-1 92

iii Given curves are,2y2=x3             ... 1y2=32x             ... 2 From these two equations we get232x=x364x=x3xx2-64=0x=0, 8 , -8Substituting the value of x in 2 we get, y1=0, 16, -16x1, y1 = 0, 08, 16  or 8, -16 Differentiating (1) w.r.t. x,4y dydx=3x2dydx=3x24y               ...3Differenntiating (2) w.r.t. x,2ydydx=32dydx=16y               ...4Case - 1: x, y = 0, 0From 3 we have, m1=00 We cannot determine θ in this case.Case - 2x, y = 8, 16From 3 we have, m1=19264=3From 4 we have, m2=1616=1Now,tan θ=m1-m21+m1m2=3-11+3=24=12θ=tan-1 12Case- 3x1, y1 8, -16From 3 we have, m1=192-64=-3From 4 we have, m2=16-16=-1Now, tan θ=m1-m21+m1m2=-3+11+3=24=12θ=tan-1 12

 iv Given curves are,x2+y2-4x-1=0  ... 1x2+y2-2y-9=0 ... 2From (3)  we getx2+y2=4x+1Substituting this in (2),4x+1-2y-9=04x-2y=82x-y=4y=2x-4      ... 3Substituting this in (1),x2+2x-42-4x-1=0x2+4x2+16-16x-4x-1=05x2-20x+15=0x2-4x+3=0x-3x-1=0x=3 or x=1Substituting the values of x in 3, we get,y=2 or y=-2 x, y = 3, 21, -2Differentiating (1) w.r.t. x,2x+2y dydx-4=0dydx=4-2x2y=2-xy         ... 4Differenntiating (2) w.r.tx,2x+2y dydx-2dydx=0dydx2y-2=-2xdydx=2x2-2y=x1-y       ... 5Case - 1: x, y = 3, 2From 4, we get, m1=2-32=-12From 5, we get, m2=31-2=-3Now,tan θ=m1-m21+m1m2=-12+31+32=1θ=tan-1 1=π4Case - 2:  x, y = 1, - 2From 4, we get, m1=2-1-2=-12From 5, we get, m2=11+2=13Now,tan θ=m1-m21+m1m2=-12-131-16=1θ=tan-1 1=π4

v  Given curves are,x2a2+y2b2=1 ... 1x2+y2=ab ... 2Multiplying (2) by 1a2,x2a2+y2a2=ba ... 3Subtracting (1) from (3), we gety2a2- y2b2=ba-1y2b2-a2a2b2=b-aay2=b-aa×a2b2b+ab-a=ab2b+ay=±bab+aSubstituting this in (3),x2a2+ab2b+aa2=baa+bx2+ab2=ab2+a2bx2=a2ba+bx=±aba+bx, y = ±aba+b, ±bab+aNow, x, y = aba+b, bab+aDifferentiating (1) w.r.t. x, we get,2xa2+2yb2dydx=0dydx=-xb2a2ym1=-ab2ba+ba2bab+a =-bbaaDifferenntiating (2) w.r.tx, we get,2x+2ydydx=0dydx=-xym2=-aba+b bab+a=-abbaWe have,tan θ=m1-m21+m1m2=-bbaa+abba1+bbaaabba=-b2ab+a2aba2ba2b+ab2a2b=aba+ba-ba2b×a2baba+b=a-babθ=tan-1 a-babSimilarly,  we can prove that θ=tan-1 a-bab for all possibilities of x, y

vi Given curves are,x2+4y2=8 ... 1x2-2y2=2 ... 2From (1) and (2) we get 6y2=6y=1 or y1=-1Substituting the values of y in 1x=2, -2 or x=2, -2 So, x, y=2, 12, -1-2, 1-2, -1Differenntiating (1) w.r.t. x,2x+8y dydx=0dydx=-x4y    ...3Differenntiating (2) w.r.t. x,2x-4y dydx=0dydx=x2y     ...4   Case -1:  x, y = 2, 1 From 3,  we get, m1=-12From 4,  we get, m2=1We have,tan θ=m1-m21+m1m2=-12-11-12=3θ=tan-1 3Case -2:  x, y = 2, -1 From 3,  we get, m1=12From 4,  we get, m2=-1We have,tan θ=m1-m21+m1m2=12+11-12=3θ=tan-1 3Case -3 x, y-2, 1 From 3,  we get, m1=12From 4,  we get, m2=-1We have,tan θ=m1-m21+m1m2=12+11-12=3θ=tan-1 3Case -4 x, y= -2, -1 From 3,  we get, m1=-12From 4,  we get, m2=1We have,tan θ=m1-m21+m1m2=-12-11-12=3θ=tan-1 3

vii Given curves are,x2=27y       ... 1y2=8x          ... 2From (2)  we getx=y28 Substituting this in (1),y282=27yy4=1728yy y3-123=0y=0 or y=12Substituting the values of y in (2), we get, x=0 or x=18x, y = 0, 018, 12Differentiating (1) w.r.t. x,2x=27dydxdydx=2x27          ... 3Differenntiating (2) w.r.tx,2y dydx=8dydx=4y          ... 4Case - 1: x, y = 0, 0From 4 we have, m2 is undefined We cannot find θCase - 2:  x, y = 18, 12From 3 we have, m1=3627=43From 4 we have, m2=412=13Now, tan θ=m1-m21+m1m2=43-131+49=913θ=tan-1 913

viii Given curves are,x2+y2=2x                     ... 1y2=x                       ... 2From these two equations we get x2+x=2xx2-x=0x x-1=0x=0 or x=1Substituting the values of x in 2 we get,y=0 or y=±1          ∴ x,y =0, 01, 11, -1Differenntiating (1) w.r.t. x,we get,2x+2ydydx=2dydx=1-xy                      ... 3Differenntiating (2) w.r.t. x,we get,2y dydx=1dydx=12y                      ... 4Case -1: x, y  = 0, 0From 3 we get, m1 is undefined. We can not find θCase -2: Let x, y 1, 1From 3 we get, m1=0From 4 we get, m2=12Now, tan θ=m1-m21+m1m2=0-121+0=12θ=tan-112Case -3: Let x, y 1, -1From 3 we get, m1=0From 4 we get, m2=-12Now, tan θ=m1-m21+m1m2=0+121=12θ=tan-112

ix Given curves are,y=4-x2       .....1y=x2             .....2From 1 and (2), we get4-x2=x2 2x2=4x2=2x=±2Substituting the values of x in (2), we get, y=2x, y=2,2-2,2Differentiating (1) w.r.t. x,dydx=-2x          .....3Differentiating (2) w.r.tx,dydx=2x          .....4Case 1: x, y=2, 2From 3, we have, m1=-22From 4 we have, m2=22Now, tanθ=m1-m21+m1m2=-22-221-8=427θ=tan-1427Case 1: x, y=-2, 2From 3, we have, m1=22From 4 we have, m2=-22Now, tanθ=m1-m21+m1m2=22+221-8=427θ=tan-1 427

Page No 16.40:

Question 2:

Show that the following set of curves intersect orthogonally.

(i) y = x3 and 6y = 7 − x2
(ii) x3 − 3xy2 = −2 and 3x2yy3 = 2
(iii) x2 + 4y2 = 8 and x2 − 2y2 = 4

Answer:

i  y=x3... 16y=7-x2 ... 2From (1)  and (2) we get6x3=7-x26x3+x2-7=0x=1 satisfies this.Dividing this by x-1 ,we get6x2+7x+7=0, Discriminant = 72-467=-119<0So, this has no real roots.When x=1, y=x3=1 (From (1))So, x, y = 1, 1Differentiating (1) w.r.t. x,dydx=3x2m1=dydx1, 1=3Differenntiating (2) w.r.tx,6dxdx=-2xdxdx=-2x6=-x3m2=dydx1, 1=-13Now, m1×m2=3×-13m1×m2=-1Since, m1×m2 = -1
So, the given curves intersect orthogonally.

ii Let the given curves intersect at x1, y1x3-3xy2=-2... 13x2y-y3=2 ... 2Differentiating (1) w.r.t. x,3x2-3y2-6xydydx=0dydx=3x2-3y26xy=x2-y22xym1=dydxx1, y1=x12-y122x1y1Differenntiating (2) w.r.tx,3x2dydx+6xy-3y2dydx=0dydx3x2-3y2=-6xydydx=-6xy3x2-3y2=-2xyx2-y2m2=dydxx1, y1=-2x1y1x12-y12Now, m1×m2=x12-y122x1y1×-2x1y1x12-y12 m1×m2=-1Since, m1×m2=-1
So, the given curves intersect orthogonally.

iii x2+4y2=8 ... 1x2-2y2=4 ... 2From (1) and (2) we get6y2=4y2=23y=23 or  y=-23From (1),x2+83=8x2=163x=±43So, x, y=43, 2343, -23,-43, 23-43, -23Consider point x1, y1=43, 23Differentiating (1) w.r.t. x,2x+8ydydx=0dydx=-x4ym1=dydx43, 23=-43423=-12Differenntiating (2) w.r.tx,2x-4ydydx=0dydx=x2ym2=dydx43, 23=43223=2Now, m1×m2=-12×2 m1×m2=-1Since, m1×m2=-1Hence,, the curves are orthogonal at 43, 23.Similarly, we can see that the curves are orthogonal in each possibility of x1, y1.

Page No 16.40:

Question 3:

Show that the following curves intersect orthogonally at the indicated points:

(i) x2 = 4y and 4y + x2 = 8 at (2, 1)
(ii) x2 = y and x3 + 6y = 7 at (1, 1)
(iii) y2 = 8x and 2x2y2 = 10 at 1, 22

Answer:

i x2=4y... 14y+x2=8 ... 2Given point is 2, 1Differentiating (1) w.r.t. x,2x=4dydxdydx=x2m1=dydx2,1=22=1Differenntiating (2) w.r.tx,4dydx+2x=0dydx=-x2m2=dydx2,1=-22=-1Since, m1×m2=-1
Hence,  the given curves intersect orthogonally at the given point.

ii x2=y... 1x3+6y=7 ... 2Given point is 1, 1Differentiating (1) w.r.t. x,2x=dydxm1=dydx1,1=21=2Differenntiating (2) w.r.tx,3x2+6dydx=0dydx=-x22m2=dydx1,1=-12Since, m1×m2=-1
Hence,  the given curves intersect orthogonally at the given point.

iiiy2=8x... 12x2+y2=10 ... 2Given point is 1, 22Differentiating (1) w.r.t. x,2ydydx=8dydx=4ym1=dydx1, 22=422=2Differenntiating (2) w.r.tx,4x+2ydydx=0dydx=-2xym2=dydx1, 22=-222=-12Since, m1×m2=-1
Hence,  the given curves intersect orthogonally at the given point.

Page No 16.40:

Question 4:

Show that the curves 4x = y2 and 4xy = k cut at right angles, if k2 = 512.

Answer:

Given:4x=y2   ...14xy=k   ...2From (1) and (2), we gety3=ky=k13From (1), we get4x=k23x=k234On differentiating (1) w.r.t. x, we get4=2ydydx dydx=2ym1=dydxk234, k13=2k13=2k-13On differentiating (2) w.r.tx, we get4xdydx+4y=0dydx=-yxm2=dydxk234, k13=-k13k234=-4k-13It is given that the curves intersect at right angles.m1×m2=-12k-13×-4k-13=-18k-23=1k-23=18k23=8Cubing on both sides, we getk2=512

Page No 16.40:

Question 5:

Show that the curves 2x = y2 and 2xy = k cut at right angles, if k2 = 8.

Answer:

Given:2x=y2     ...12xy=k     ...2From (1)  and (2), we gety3=ky=k13From (1), we get2x=k2 3x=k232On differentiating (1) w.r.t. x, we get2=2ydydx dydx=1ym1=dydxk232, k13=1k13=k-13On differentiating (2) w.r.tx, we get2xdydx+2y=0dydx=-yxm2=dydxk232, k13=-k13k232=-2k-13It is given that the curves intersect orthogonally.m1×m2=-1k-13×-2k-13=-12k-23=1k-23=12k23=2Cubing on both sides, we getk2=8

Page No 16.40:

Question 6:

Prove that the curves xy = 4 and x2 + y2 = 8 touch each other.                                                                                        [NCERT EXEMPLAR]

Answer:

Given:xy=4              .....1x2+y2=8      .....2From 1, we getx=4ySubstituting x=4y in 2, we get4y2+y2=816y2+y2=816+y4=8y2y4-8y2+16=0y2-42=0y2-4=0y2=4y=±2Substituting y=±2, we getx=±2So, the given curves touch each other at two points 2,2 and -2,-2.

Page No 16.40:

Question 7:

Prove that the curves y2 = 4x and x2 + y2 - 6x + 1 = 0 touch each other at the point (1, 2).                                         [NCERT EXEMPLAR]

Answer:

Given:y2=4x                          .....1 andx2+y2-6x+1=0      .....2From 1 and 2, we getx2+4x-6x+1=0x2-2x+1=0x-12=0x-1=0x=1Substititing x=1 in 1, we gety2=4y=±2So, the two given curves touch each other at two points 1,2 and 1,-2.



Page No 16.41:

Question 8:

Find the condition for the following set of curves to intersect orthogonally:

(i) x2a2-y2b2=1 and xy=c2
(ii) x2a2+y2b2=1 and x2A2-y2B2=1

Answer:


i x2a2-y2b2=1   ...1xy=c2 ...2Let the curves intersect orthogonally at x1, y1.On differentiating (1) on both sides w.r.t. x, we get2xa2-2yb2dydx=0dydx=xb2a2ym1=dydxx1, y1=x1b2a2y1On differentiating (2) on both sides w.r.t. x, we getxdydx+y=0dydx=-yxm2=dydxx1, y1=-y1x1It is given that the curves intersect orhtogonally at x1, y1.m1×m2=-1x1b2a2y1×-y1x1=-1a2=b2

(ii) The condition for the curves ax2+by2=1 and a'x2+b'y2=1 to intersect orthogonally is given below:
1a-1b=1a'-1b'So, the condition for the curves  x2a2+y2b2=1 and x2A2-y2B2=1 to intersect orthogonally is11a2-11b2=11A2-1-1B2a2-b2=A2+B2

Page No 16.41:

Question 9:

Show that the curves x2a2+λ1+y2b2+λ1=1 and x2a2+λ2+y2b2+λ2=1intersect at right angles.

Answer:

We have, x2a2+λ1+y2b2+λ1=1    ...1and x2a2+λ2+y2b2+λ2=1           ...2Now we can find the slope of both the curve by differentiating w.r.t x2xa2+λ1+2ydydxb2+λ1=0 and 2xa2+λ2+2ydydxb2+λ2=0dydx=-xy×b2+λ1a2+λ1 and dydx=-xy×b2+λ2a2+λ2m1=-xy×b2+λ1a2+λ1 and m2=-xy×b2+λ2a2+λ2Subtracting 2 from 1, we get,x21a2+λ1-1a2+λ2+y21b2+λ1-1b2+λ2=0x2y2=λ2-λ1b2+λ1b2+λ2×1λ1-λ2a2+λ1a2+λ2Now,m1××m2=x2y2×b2+λ1a2+λ1×b2+λ2a2+λ2                 =λ2-λ1b2+λ1b2+λ2×a2+λ1a2+λ2λ1-λ2×b2+λ1a2+λ1×b2+λ2a2+λ2                 =-1hence, 1 and 2 cuts orthogonally.

Page No 16.41:

Question 10:

If the straight line xcosα + ysinα = p touches the curve x2a2-y2b2=1, then prove that a2cos2α - b2sin2α = p2.

Answer:

We have,xcosα+ysinα=p          .....ix2a2-y2b2=1                 .....iiAs, the straight line i touches the curve ii.So, the straight line i is tangent to the curve ii.Also, the slope of the straight line, m=-cosαsinαAnd, the slope of the tangent to the curve=dydx=b2a2×xySo, b2a2×xy=-cosαsinαxb2sinα=-ya2cosαx=-ya2cosαb2sinα             .....iiiSubstituting the value of x in i, we getxcosα+ysinα=p-ya2cos2αb2sinα+ysinα=p-ya2cos2α+yb2sinαb2sinα=py-a2cos2α+b2sinα=pb2sinαy=pb2sinαb2sinα-a2cos2αSo, from iii, we getx=-ya2cosαb2sinα=-ya2cosαb2sinα=-a2cosαb2sinα×pb2sinαb2sinα-a2cos2αx=-pa2cosαb2sinα-a2cos2αSubstituting the values x and y in ii, we getx2a2-y2b2=11a2×-pa2cosαb2sinα-a2cos2α2-1b2×pb2sinαb2sinα-a2cos2α2=1p2a2cos2αb2sinα-a2cos2α2-p2b2sin2αb2sinα-a2cos2α2=1p2a2cos2α-p2b2sin2αb2sinα-a2cos2α2=1p2a2cos2α-b2sin2αb2sinα-a2cos2α2=1p2a2cos2α-b2sin2αa2cos2α-b2sin2α2=1p2a2cos2α-b2sin2α=1 p2=a2cos2α-b2sin2α

Page No 16.41:

Question 1:

Find the point on the curve y = x2 − 2x + 3, where the tangent is parallel to x-axis.

Answer:

The slope of the x-axis is 0.
Now, let (x1, y1) be the required point.
Here,
Since, the point lies on the curve.Hence, y1=x12-2x1+3            ... 1Now, y=x2-2x+3dydx=2x-2Slope of the tangent at x, y=dydxx1, y1=2x1-2Given:Slope of the tangent at x1, y1=Slope of the x-axis=2x1-2=0x1=1andy1=1-2+3=2         [From (1)]Required point=x1, y1=1, 2

Page No 16.41:

Question 2:

Find the slope of the tangent to the curve x = t2 + 3t − 8, y = 2t2 − 2t − 5 at t = 2.

Answer:

Here,x=t2+3t-8 and y=2t2-2t-5dxdt=2t+3 and dydt=4t-2dydx=dydtdxdt=4t-22t+3Now,Slope of the tangent=dydxt=2=8-24+3=67

Page No 16.41:

Question 3:

If the tangent line at a point (x, y) on the curve y = f(x) is parallel to x-axis, then write the value of dydx.

Answer:

The slope of the x-axis is 0.
Also, the tangent at a point (x, y) on the curve y = f(x) is parallel to the x-axis.
∴ Slope of the tangent dydx= Slope of the x-axis = 0

Page No 16.41:

Question 4:

Write the value of dydx, if the normal to the curve y = f(x) at (x, y) is parallel to y-axis.

Answer:

The slope of the y-axis is .
Also, the normal at (x, y) on the curve y = f(x) is parallel to the y-axis.
∴ Slope of the normal = Slope of the y-axis =
⇒ dydx=Slope of the tangent=-1Slope of the normal=-1=0

Page No 16.41:

Question 5:

If the tangent to a curve at a point (x, y) is equally inclined to the coordinates axes then write the value of dydx.

Answer:

Because the tangent to the curve at (x, y) is equally inclined to the coordinate axes, the angle made by the tangent with the axes can be ±45°.
dydx=Slope of the tangent=tan ±45=±1

Page No 16.41:

Question 6:

If the tangent line at a point (x, y) on the curve y = f(x) is parallel to y-axis, find the value of dxdy.

Answer:

Slope of the y-axis is .
Also, the tangent at (x, y) on the curve y = f(x) is parallel to the y-axis,
∴ Slope of the tangent, dydx = Slope of the y-axis =
dxdy=1dydx=1=0

Page No 16.41:

Question 7:

Find the slope of the normal at the point 't' on the curve x=1t, y=t.

Answer:

Here,x=1t and y=tdxdt=-1t2and dydt=1dydx=dydtdxdt=1-1t2=-t2Now,Slope of the tangent=dydx=-t2Slope of the normal=-1 Slope of the tangent=-1-t2=1t2 

Page No 16.41:

Question 8:

Write the coordinates of the point on the curve y2 = x where the tangent line makes an angle π4 with x-axis.

Answer:

Let the required point be (x1, y1).
The tangent makes an angle of 45o with the x-axis.
∴ Slope of the tangent = tan 45o = 1

Since, the point lies on the curve.Hence, y12=x1Now, y2=x2ydydx=1dydx=12ySlope of the tangent = dydxx1, y1=12y1Given:12y1=12y1=1y1=12Now,x1=y12=122=14x1, y1=14, 12

Page No 16.41:

Question 9:

Write the angle made by the tangent to the curve x = et cos t, y = et sin t at t=π4 with the x-axis.

Answer:

Here,x=et cos t and y=et sin tdxdt=et cos t-et sin t  anddydt=et sin t+et cos tdydx=dydtdxdt=et sin t+et cos tet cos t-et sin t=sin t+cos tcos t-sin tNow,Slope of the tangent =dydxt=π4=sin π4+cos π4cos π4-sin π4=12+1212-12=220=∞Let θ be the angle made by the tangent with the x-axis.tan θ=∞θ=π2



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Question 10:

Write the equation of the normal to the curve y = x + sin x cos x at x=π2.

Answer:

Here,y=x+sin x cos xOn differentiating both sides w.r.t. x, we getdydx=1+cos2 x-sin2xNow,Slope of the tangent = dydxx=π2= 1+cos2π2-sin2π2= 1-1=0When x=π2y=π2+sin π2cosπ2=π2x1, y1=π2, π2Equation of the normal=y-y1=-1Slope of the tangentx-x1y-π2=-10x-π2x=π22x=π

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Question 11:

Find the coordinates of the point on the curve y2 = 3 − 4x where tangent is parallel to the line 2x + y − 2 = 0.

Answer:

Let (x1, y1) be the required point.
Slope of the given line = -2

Since, the point lies on the curve.Hence, y12=3-4x1       ...1Now, y2=3-4x2ydydx=-4dydx=-42y=-2ySlope of the tangent = dydxx1, y1=-2y1Given:Slope of the tangent = Slope of the line-2y1=-2y1=1From (1), we get1=3-4x1-2=-4x1x1=12x1, y1=12, 1

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Question 12:

Write the equation on the tangent to the curve y = x2x + 2 at the point where it crosses the y-axis.

Answer:

When the curve crosses the y-axis, the point on the curve is of the form (0, y).
Here,
y=x2-x+2y=0-0+2=2So, the point where the curve crosses the y-axis is (0, 2).Now, y=x2-x+2dydx=2x-1Slope of the tangent, m = dydx0, 2=20-1=-1x1, y1=0, 2andEquation of tangent=y-y1=mx-x1y-2=-1x-0y-2=-xx+y-2=0

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Question 13:

Write the angle between the curves y2 = 4x and x2 = 2y − 3 at the point (1, 2).

Answer:

Given:y2=4x        ...1x2=2y-3  ...2On differentiating (1) w.r.t. x, we get2ydydx=4dydx=2ym1=dydx1, 2=22=1On differentiating (2) w.r.t. x, we get2x=2dydxdydx=xm2=dydx1, 2=1Thus, we gettan θ=m1-m21+m1m2tan θ=1-11+1tan θ=0θ=0o

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Question 14:

Write the angle between the curves y = ex and y = ex at their point of intersections.

Answer:

Given:y=e-x    ...1y=ex       ...2On substituting the value of y in (1), we gete-x=exx=0andy=1 [From (2)]On differentiating (1) w.r.t. x,we getdydx=-e-xm1=dydx0, 1=-1On differentiating (2) w.r.tx, we getdydx=exm2=dydx0, 1=1m1×m2=-1Since the multiplication of the slopes is -1 so the slopes are perpendicular to each other.Required angle=π2

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Question 15:

Write the slope of the normal to the curve y=1xat the point 3, 13.

Answer:

Given:y=1xOn differentiating both sides w.r.t. x, we getdydx=-1x2Now,Slope of the tangent = dydx3, 13=-19Slope of the normal = -1Slope of tangent=-1-19= 9

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Question 16:

Write the coordinates of the point at which the tangent to the curve y = 2x2x + 1 is parallel to the line y = 3x + 9.

Answer:

Let (x1, y1) be the required point.
Slope of the given line = 3
Since, the point lies on the curve.Hence, y1=2x12-x1+1       ...1 Now, y=2x2-x+1 dydx=4x-1Now,Slope of the tangent = dydxx1, y14x1-1Given:Slope of the tangent = Slope of line4x1-1=3x1=1From (1), we gety1=2-1+1=2x1, y1=1, 2

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Question 17:

Write the equation of the normal to the curve y = cos x at (0, 1).

Answer:

Given:y=cos xOn differentiating both sides w.r.t. x, we getdydx=-sin xNow,Slope of the tangent=dydx0, 1=-sin 0=0andEquation of the normal=y-y1=-1mx-x1y-1=-10x-0x=0

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Question 18:

Write the equation of the tangent drawn to the curve y=sinx at the point (0,0).

Answer:

We have,

y=sinx

dydx=cosx

Slope at (0, 0) = mdydxx=0=cos0=1

So, the equation of the tangent at (0,0) is given by,

y = mx

Putting m = 1, we get

The equation of the tangent is y = x.

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Question 1:

The equation to the normal to the curve y = sin x at (0, 0) is

(a) x = 0
(b) y = 0
(c) x + y = 0
(d) xy = 0

Answer:

(c) x + y = 0

Given:y=sin xOn differentiating both sides w.r.t. x, we getdydx=cos xSlope of the tangent = dydx0, 0= cos 0 =1Slope of the normalm = -11=-1Given:x1, y1=0, 0Equation of the normal=y-y1=mx-x1y-0=-1x-0y=-xx+y=0

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Question 2:

The equation of the normal to the curve y = x + sin x cos x at x = π/2 is

(a) x = 2
(b) x = π
(c) x + π = 0
(d) 2x = π

Answer:

(d) 2x = π

Given:y=x+sin x cosxOn differentiating both sides w.r.t. x, we getdydx=1+cos2 x-sin2 xSlope of the tangent=dydxx=π2=1+cos 2π2-sin2π2=1-1=0Slope of the normal, m=-10When x=π2, y=π2+cos π2 sin π2=π2Now, x1, y1=π2, π2Equation of the normal=y-y1=mx-x1y-π2=-10x-π2x=π22x=π

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Question 3:

The equation of the normal to the curve y = x(2 − x) at the point (2, 0) is

(a) x − 2y = 2
(b) x − 2y + 2 = 0
(c) 2xy = 4
(d) 2x + y − 4 = 0

Answer:

(a) x − 2y = 2

Here,y=x2-x=2x-x2dydx=2-2xSlope of the tangent=dydx2, 0=2-4=-2Slope of the normal, m=-1-2=12Given:x1, y1=2, 0Equation of the normal=y-y1=mx-x1y-0=12x-22y=x-2x-2y=2

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Question 4:

The point on the curve y2 = x where tangent makes 45° angle with x-axis is

(a) (1/2, 1/4)
(b) (1/4, 1/2)
(c) (4, 2)
(d) (1, 1)

Answer:

(b) (1/4, 1/2)

Let the required point be (x1, y1).
The tangent makes an angle of 45o with the x-axis.
∴ Slope of the tangent = tan 45o = 1

Since, the point lies on the curve.Hence, y12=x1Now, y2=x2ydydx=1dydx=12ySlope of the tangent = dydxx1, y1=12y1Given:12y1=12y1=1y1=12Now,x1=y12=122=14x1, y1=14, 12

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Question 5:

If the tangent to the curve x = a t2, y = 2 at is perpendicular to x-axis, then its point of contact is

(a) (a, a)
(b) (0, a)
(c) (0, 0)
(d) (a, 0)

Answer:

(c) (0, 0)

Let the required point be (x1, y1).

Since, the point lies on the curve.Hence, x1=at2 and y1=2atNow, x=at2 and  y=2atdxdt=2at and  dydt=2adydx=dydtdxdt=2a2at=1t=2aySlope of the tangent = dydxx1, y1=2ay1It is given that the tangent is perpendicular to the y-axis.It means that it is parallel to the x-axis.∴ Slope of the tangent = Slope of the  x-axis2ay1=0a=0Nowx1=at2=0 and y1=2at=0x1, y1=0, 0

 

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Question 6:

The point on the curve y = x2 − 3x + 2 where tangent is perpendicular to y = x is

(a) (0, 2)
(b) (1, 0)
(c) (−1, 6)
(d) (2, −2)

Answer:

(b) (1, 0)

y = x
dydx=1

Let x1, y1 be the required point.Since, the point lies on the curve,Hence, y1=x12-3x1+2Now, y=x2-3x+2dydx=2x-3Slope of the tangent=dydxx1, y1=2x1-3
The tangent is perpendicular to this line.
∴Slope of the tangent = -1Slope of the line=-11=-1

Now,
2x1-3=-12x1=2x1=1andy1=x12-3x1+2=1-3+2=0x1, y1=1, 0

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Question 7:

The point on the curve y2 = x where tangent makes 45° angle with x-axis is

(a) (1/2, 1/4)
(b) (1/4, 1/2)
(c) (4, 2)
(d) (1, 1)

Answer:

(b) (1/4, 1/2)

Let the required point be (x1, y1).
The tangent makes an angle of 45o with the x-axis.
∴ Slope of the tangent = tan 45o = 1

Since, the point lies on the curve.Hence, y12=x1Now, y2=x2ydydx=1dydx=12ySlope of the tangent = dydxx1, y1=12y1Given:12y1=12y1=1y1=12Now,x1=y12=122=14x1, y1=14, 12

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Question 8:

The point at the curve y = 12xx2 where the slope of the tangent is zero will be

(a) (0, 0)
(b) (2, 16)
(c) (3, 9)
(d) none of these

Answer:

(d) none of these

Let x1, y1 be the required point.Since, the point lies on the curve,Hence, y1=12x1-x12Now,y=12x-x2dydx=12-2xSlope of the tangent = dydxx1, y1=12-2x1Given:Slope of the tangent=012-2x1=0x1=6Now,y1=12x1-x12=72-36=36x1, y1=6, 36

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Question 9:

The angle between the curves y2 = x and x2 = y at (1, 1) is

(a) tan-1 43
(b) tan-1 34
(c) 90°
(d) 45°

Answer:

(b) tan-1 34

Given:y2=x    ...1x2=y    ...2Point=1, 1On differentiating (1) w.r.t. x, we get2y dydx=1dydx=12ym1=12On differentiating (2) w.r.t. x, we get2x = dydxm2=21=2Now,tan θ=m1-m21+m1m2=12-21+12×2=34θ=tan-1 34



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Question 10:

The equation of the normal to the curve 3x2y2 = 8 which is parallel to x + 3y = 8 is

(a) x + 3y = 8
(b) x + 3y + 8 = 0
(c) x + 3y ± 8 = 0
(d) x + 3y = 0

Answer:

(c) x + 3y ± 8 = 0

The slope of line x + 3y = 8 is -13.

Since the normal is parallel to the given line, the equation of normal will be of the given form.x+3y=k       3x2-y2=8    Let x1, y1 be the point of intersection of the two curves.Then, x1+3y1=k         ...13x12-y12=8      ...2Now, 3x2-y2=8  On differentiating both sides w.r.t. x, we get6x-2ydydx=0dydx=6x2y=3xySlope of the tangent = dydxx1, y1=3x1y1Slope of the normalm = -13xy=-y13x1Given:Slope of the normal = Slope of the given line-y13x1=-13y1=x1       ...3From (2), we get3x12-x12=82x12=8x12=4x1=±2Case 1:When x1=2 From (3), we gety1=x1=2x1, y=2, 2From (1), we get2+32=k2+6=kk=8Equation of the normal from (1)x+3y=8x+3y-8=0Case 2:When x1=-2 From (3), we gety1=x1=-2,x1, y=-2, -2From (1), we get-2+3-2=k-2-6=kk=-8Equation of the normal from (1)x+3y=-8x+3y+8=0From both the cases, we get the equation of the normal as:x+3y±8=0

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Question 11:

The equations of tangent at those points where the curve y = x2 − 3x + 2 meets x-axis are

(a) xy + 2 = 0 = xy − 1
(b) x + y − 1 = 0 = xy − 2
(c) xy − 1 = 0 = xy
(d) xy = 0 = x + y

Answer:

(b) x + y − 1 = 0 = xy − 2

Let the tangent meet the x-axis at point (x, 0).
Now,
y=x2-3x+2dydx=2x-3The tangent passes through point (x, 0).0=x2-3x+2x-2x-1=0x=2 or x=1Case 1: When x=2:Slope of the tangent, m=dydx2, 0=4-3=1x1, y1=2, 0Equation of the tangent:y-y1=m x-x1y-0=1 x-2x-y-2=0Case 2: When x=1:Slope of the tangent, m=dydx2, 0=2-3=-1x1, y1=1, 0Equation of the tangent:y-y1=m x-x1y-0=-1 x-1x+y-1=0

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Question 12:

The slope of the tangent to the curve x = t2 + 3 t − 8, y = 2t2 − 2t − 5 at point (2, −1) is

(a) 22/7
(b) 6/7
(c) −6
(d) none of these

Answer:

(b) 6/7

x=t2+3t-8 and y=2t2-2t-5dxdt=2t+3 and dydt=4t-2dydx=dydtdxdt=4t-22t+3The given point is (2, -1). x=2 and y=-1Now,t2+3t-8=2 and 2t2-2t-5=-1Let us solve one of these to get the value of t.t2+3t-10=0 and 2t2-2t-4=0t+5t-2=0 and 2t+2t-2=0t=-5 or t= 2 and t=-1 or t=2These two have t=2 as a common solution.Slope of the tangent = dydxt=2= 8-24+3=67

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Question 13:

At what point the slope of the tangent to the curve x2 + y2 − 2x − 3 = 0 is zero

(a) (3, 0), (−1, 0)
(b) (3, 0), (1, 2)
(c) (−1, 0), (1, 2)
(d) (1, 2), (1, −2)

Answer:

(d) (1, 2), (1, −2)

Let (x1, y1) be the required point.

Since, the point lie on the curve.Hence, x12+y12-2x1-3=0   ...1Now, x2+y2-2x-3=0 2x+2y dydx-2=0dydx=2-2x2y=1-xyNow,Slope of the tangent = dydxx1, y11-x1y1Slope of the tangent=0    (Given)1-x1y1=01-x1=0x1=1From (1), we getx12+y12-2x1-3=01+y12-2-3=0y12-4=0y1=±2So, the points are 1, 2 and 1, -2.

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Question 14:

The angle of intersection of the curves xy = a2 and x2y2 = 2a2 is

(a) 0°
(b) 45°
(c) 90°
(d) none of these

Answer:

(c) 90°

Given:xy=a2             ...1x2-y2=2a2    ...2Let x1, y1 be the point of intersection.On differentiating (1) w.r.t. x, we getxdydx+y=0dydx=-yxm1=dydxx1, y1=-y1x1On differentiating (2) w.r.t. x, we get2x -2y dydx=0dydx=xym2=dydxx1, y1=x1y1Now, m1×m2=-y1x1×x1y1 m1×m2=-1m1×m =-1So, the angle between the curves is 90°.

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Question 15:

If the curve ay + x2 = 7 and x3 = y cut orthogonally at (1, 1), then a is equal to

(a) 1
(b) −6
(c) 6
(d) 0

Answer:

(c) 6

Given:ay+x2=7  ...1x3=y           ...2Point=1, 1On differentiating (1) w.r.t. x, we getadydx+2x=0dydx=-2xam1=dydx1, 1=-2aAgain, on differentiating (2) w.r.t. x, we get3x2=dydxm2=dydx1, 1=3It is given that the curves are orthogonal at the given point.m1×m2=-1-2a×3=-1a=6

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Question 16:

If the line y = x touches the curve y = x2 + bx + c at a point (1, 1) then

(a) b = 1, c = 2
(b) b = −1, c = 1
(c) b = 2, c = 1
(d) b = −2, c = 1

Answer:

(b) b = −1, c = 1

We can find the slope of the line by differentiating w.r.t. x.
Slope of the given line = 1
Now,
y=x2+bx+c   ...1dydx=2x+bSlope of the tangent= dydx1, 1=2+bGiven:Slope of the tangent = 12+b=1b=-1On substituting b=-1, x=1 and y=1 in (1), we get1=1-1+cc=1

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Question 17:

The slope of the tangent to the curve x = 3t2 + 1, y = t3 −1 at x = 1 is

(a) 1/2
(b) 0
(c) −2
(d) ∞

Answer:

(b) 0

Given:x=3t2+1 y=t3-1x=1Now,3t2+1=13t2=0t=0dxdt=6t and dydt=3t2dydx=dydtdxdt=3t26t=t2Thus, we getSlope of the tangent=dydxt=0=02=0

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Question 18:

The curves y = aex and y = bex cut orthogonally, if

(a) a = b
(b) a = −b
(c) ab = 1
(d) ab = 2

Answer:

(c) ab = 1

Given:y=aex     ...1y=be-x   ...2Let the point of intersection of these two curves be x1, y1.Now,On differentiating (1) w.r.t. x, we getdydx=aexm1=dydx x1, y1 =aex1Again, on differentiating (2) w.r.t. x, we getdydx=-be-xm2=dydx x1, y1=-be-x1It is given that the curves cut orthogonally.m1×m2=-1aex1×-be-x1=-1ab=1

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Question 19:

The equation of the normal to the curve x = a cos3 θ, y = a sin3 θ at the point θ = π/4 is

(a) x = 0
(b) y = 0
(c) c = y
(d) x + y = a

Answer:

(c) x = y

Here,x=a cos3 θ and y=a sin3 θdxdθ=-3a cos2 θ sin θ  and dydθ=3a sin2 θ cos θdydx=dydθdxdθ=3a sin2 θ cos θ-3a cos2 θ sin θ=-tan θNow,Slope of the tangent = dydxθ=π4=-tan π4=-1x1, y1=a cos3 π4, a sin3 π4=a22, a22Equation of the normal=y-y1=-1m x-x1y-a22=1 x-a22x=y

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Question 20:

If the curves y = 2 ex and y = aex intersect orthogonally, then a =

(a) 1/2
(b) −1/2
(c) 2
(d) 2e2

Answer:

(a) 1/2

Given:y=2ex     ...1y=ae-x   ...2Let the point of intersection of these curves be x1, y1.On differentiating (1) w.r.t. x, we getdydx=2exm1=dydx x1, y1=2ex1Now, on differentiating (2) w.r.t. x, we getdydx=-ae-xm2=dydx x1, y1=-ae-x1It is given that the curves are orthogonal.m1×m2=-12ex1×-ae-x1=-12a=1a=12

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Question 21:

The point on the curve y = 6xx2 at which the tangent to the curve is inclined at π/4 to the line x + y = 0 is

(a) (−3, −27)
(b) (3, 9)
(c) (7/2, 35/4)
(d) (0, 0)

Answer:

(b) (3, 9)

Let (x1, y1) be the point where the given curve intersect the given line at the given angle.
Since, the point lie on the curve.Hence, y1=6x1-x12Now, y=6x-x2dydx=6-2xm1=6-2x1andx+y=01+dydx=0 dydx=-1m2=-1It is given that the angle between them is π4.tan θ=m1-m21+m1m2tan π4=6-2x1+11-6+2x11=7-2x12x1-57-2x12x1-5=±17-2x12x1-5=1 or 7-2x12x1-5=-17-2x1=2x1-5 or 7-2x1=-2x1+54x1=12 or 2 = 0 (It is not true.)x1=3andy1=6x1-x12=18-9=9x1, y1=3, 9

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Question 22:

The angle of intersection of the parabolas y2 = 4 ax and x2 = 4ay at the origin is

(a) π/6
(b) π/3
(c) π/2
(d) π/4

Answer:

(c) π/2

Given:y2=4ax  ...1x2=4ay  ...2Point = 0, 0On differentiating (1) w.r.t. x, we get2y dydx=4adydx=2aym1=Now, on differentiating (2) w.r.t. x, we get2x=4adydxdydx=x2a=0tan θ=m1-m21+m1m2=1+0=θ=tan-1=π2

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Question 23:

The angle of intersection of the curves y = 2 sin2 x and y = cos 2 x at x=π6 is

(a) π/4
(b) π/2
(c) π/3
(d) none of these

Answer:

(c) π/3

Given:x=π6Now,y=2sin2xdydx=4sinx cosxdydx=2 sin 2xm1=dydxx=π6=2×32=3Also,y=cos 2xdydx=-2 sin2xm2=dydxx=π6=-2 ×32=-3tan θ=m1-m21+m1m2=3+31-33=23-2=3θ=tan-13=π3

Page No 16.43:

Question 24:

Any tangent to the curve y = 2x7 + 3x + 5

(a) is parallel to x-axis
(b) is parallel to y-axis
(c) makes an acute angle with x-axis
(d) makes an obtuse angle with x-axis

Answer:

(c) makes an acute angle with x-axis

We have, y = 2x7 + 3x + 5
dydx=14x6+3dydx>3    x6 is always positive for any real value of xdydx>0So, tanθ>0Hence, θ lies in first quadrant.Thus, the tangent to the curve makes an acute angle with x-axis

Page No 16.43:

Question 25:

The point on the curve 9y2 = x3, where the normal to the curve makes equal intercepts with the axes is

(a) 4, 83
(b) -4, 83
(c) 4, -83
(d) none of these

Answer:

(a) 4, 83 and (c) 4, -83

Let (x1, y1) be the required point.
Since, x1,y1 lies on the given curve9y12=x13               ...1Now, 9y2=x3 18y dydx=3x2dydx=3x218y=x26ySlope of the tangent = dydxx1, y1=x126y1Slope of the normal = -1x126y1=-6y1x12It is given that the normal makes equal intercepts with the axes.∴ Slope of the normal = ±1Now,-6y1x12=±1-6y1x12=1 or -6y1x12=-1y1=-x126 or y1=x126       ...2Case 1: When y1=-x126From (1), we have9x1436=x13x14=4x13x14-4x13=0x13x1-4=0x1=0, 4Putting x1=0 in 1, we get,9y12=0y1=0Putting x1=4 in 1, we get,9y12=43y1=±83But, the line making the equal intercepts with the coordinate axes can not pass through the origin.So,  the points are 4, ±83 

Page No 16.43:

Question 26:

The slope of the tangent to the curve x = t2 + 3t − 8, y = 2t2 − 2t − 5 at the point (2, −1) is

(a) 227
(b) 67
(c) 76
(d) -67

Answer:

(b) 67

Given:
x = t2 + 3t − 8 and y = 2t2 − 2t − 5

dxdt=2t+3 and dydt=4t-2dydx=dydtdxdt=4t-22t+3The given point is (2, -1).∴ x = 2 and y = -1Now,t2+3t-8=2 and 2t2-2t-5=-1t2+3t-10=0 and t2-t-2=0t+5t-2=0 and t+1t-2=0t=-5, 2 and t=-1, 2We can see that t=2 satisfies both of these.Slope of the tangent = dydxt=28-24+3=67
Thus, the correct option is (b).

Page No 16.43:

Question 27:

The line y = mx + 1 is a tangent to the curve y2 = 4x, if the value of m is

(a) 1
(b) 2
(c) 3
(d) 12

Answer:

(a) 1
Let (x1, y1) be the required point.
The slope of the given line is m.
We have
y2=4x2y dydx=4dydx=42y=2ySlope of the tangent =dydx x1, y1=2y1Given:Slope of the tangent = mNow,2y1=m         ...1

Because the given line is a tangent to the given curve at point (x1, y1), this point lies on both the line and the curve.

y1=mx1+1 and y12=4x1x1=y1-1m and  x1=y124So, y1-1m=y124y1-12y1=y124 [From (1)]y1y1-12=y1242y12-2y1=y12y12-2y1=0y12-2y1=0y1y1-2=0y1=0, 2So, For y1=0, m=20=For y1=2, m=22=1



Page No 16.44:

Question 28:

The normal at the point (1, 1) on the curve 2y + x2 = 3 is

(a) x + y = 0
(b) xy = 0
(c) x + y + 1 = 0
(d) xy = 1

Answer:

(b) xy = 0

Given:2y+x2=32dydx+2x=0dydx=-2x2=-xSlope of the tangent = dydx1, 1=-1Slope of the normal, m = -1Slope of the tangent = -1-1=1Now,x1, y1=1, 1Equation of the normal=y-y1=m x-x1y-1=1 x-1x-y=0

Page No 16.44:

Question 29:

The normal to the curve x2 = 4y passing through (1, 2) is

(a) x + y = 3
(b) xy = 3
(c) x + y = 1
(d) xy = 1

Answer:

Disclaimer: None of the given options is correct.

Given:x2=4y2x=4dydxdydx=2x4=x2Slope of the tangent = dydx1, 2=12Slope of the normal, m = -1Slope of the tangent=-112=-2Also,x1, y1=1, 2Equation of the normal=y-y1=m x-x1y-2=-2 x-1y-2=-2x+22x+y=4



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