Rd Sharma Xi 2019 Solutions for Class 11 Humanities Math Chapter 27 Hyperbola are provided here with simple step-by-step explanations. These solutions for Hyperbola are extremely popular among Class 11 Humanities students for Math Hyperbola Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma Xi 2019 Book of Class 11 Humanities Math Chapter 27 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma Xi 2019 Solutions. All Rd Sharma Xi 2019 Solutions for class Class 11 Humanities Math are prepared by experts and are 100% accurate.

Page No 27.13:

Question 1:

The equation of the directrix of a hyperbola is xy + 3 = 0. Its focus is (−1, 1) and eccentricity 3. Find the equation of the hyperbola.

Answer:

Let be the focus and Px,y be any point on the hyperbola.
Draw PM perpendicular to the directrix.
By definition:
SP = ePM



x--12+y-12=3×x-y+32
Squaring both the sides, we get:
x+12+y-12=92x-y+32x2+2x+1+y2-2y+1=92x2+y2+9-2xy-6y+6x2x2+4x+2+2y2-4y+2=9x2+9y2+81-18xy-54y+54x7x2+7y2+50x-50y-18xy+77=0
Equation of the hyperbola:
7x2+7y2+50x-50y-18xy+77=0

Page No 27.13:

Question 2:

Find the equation of the hyperbola whose
(i) focus is (0, 3), directrix is x + y − 1 = 0 and eccentricity = 2
(ii) focus is (1, 1), directrix is 3x + 4y + 8 = 0 and eccentricity = 2
(iii) focus is (1, 1) directrix is 2x + y = 1 and eccentricity = 3
(iv) focus is (2, −1), directrix is 2x + 3y = 1 and eccentricity = 2
(v) focus is (a, 0), directrix is 2xy + a = 0 and eccentricity = 43
(vi) focus is (2, 2), directrix is x + y = 9 and eccentricity = 2.

Answer:

(i) Let be the focus and Px,y be any point on the hyperbola.
Draw PM perpendicular to the directrix.

By definition:
SP = ePM
(x-0)2+(y-3)2=2x+y-12
Squaring both the sides:
(x-0)2+(y-3)2=4x+y-122x2+y2+9-6y=2x2+y2+1+2xy-2y-2xx2+y2+4xy+2y-4x-7=0
∴ Equation of the hyperbola = x2+y2+4xy+2y-4x-7=0

(ii) Let be the focus and Px,y be any point on the hyperbola.
Draw PM perpendicular to the directrix.
By definition:
SP = ePM


(x-1)2+(y-1)2=23x+4y+85
Squaring both the sides:
(x-1)2+(y-1)2=43x+4y+852x2+1-2x+y2+1-2y=4259x2+16y2+64+24xy+64y+48x25x2+25-50x+25y2+25-50y=36x2+64y2+256+96xy+256y+192x11x2+39y2+96xy+306y+242x+206=0
∴ Equation of the hyperbola = 11x2+39y2+96xy+306y+242x+206=0

(iii) Let be the focus and Px,y be any point on the hyperbola.
Draw PM perpendicular to the directrix.
By definition:
SP = ePM

(x-1)2+(y-1)2=32x+y-15
Squaring both the sides:
(x-1)2+(y-1)2=32x+y-152x2+1-2x+y2+1-2y=354x2+y2+1+4xy-2y-4x5x2+5-10x+5y2+5-10y=12x2+3y2+3+12xy-6y-12x7x2-2y2+12xy+4y-2x-7=0
∴ Equation of the hyperbola = 7x2-2y2+12xy+4y-2x-7=0

(iv) Let be the focus and Px,y be any point on the hyperbola.
Draw PM perpendicular to the directrix.




By definition:
SP = ePM
= ePM
(x-2)2+(y+1)2=22x+3y-113
Squaring both the sides:
(x-2)2+(y+1)2=42x+3y-1132x2+4-4x+y2+1+2y=4134x2+9y2+1+12xy-6y-4x13x2+52-52x+13y2+13+26y=16x2+36y2+4+48xy-24y-16x3x2+23y2+48xy-50y+36x-61=0
∴ Equation of the hyperbola = 3x2+23y2+48xy-50y+36x-61=0

(v) Let be the focus and Px,y be any point on the hyperbola.
Draw PM perpendicular to the directrix.


By definition:
SP = ePM
(x-a)2+(y-0)2=432x-y+a5
Squaring both the sides:
(x-a)2+(y)2=1692x-y+a52x2-2ax+a2+y2=16454x2+y2+a2-4xy-2ya+4xa45x2-90ax+45a2+45y2=64x2+16y2+16a2-64xy-32ay+64ax19x2-29y2-64xy-32ay+154ax-29a2=0
∴ Equation of the hyperbola = 19x2-29y2-64xy-32ay+154ax-29a2=0

(vi) Let be the focus and Px,y be any point on the hyperbola.
Draw PM perpendicular to the directrix.
By definition:
SP = ePM

(x-2)2+(y-2)2=2x+y-92
Squaring both the sides:
(x-2)2+(y-2)2=4x+y-922x2-4x+4+y2-4y+4=2x2+y2+81+2xy-18y-18xx2-4x+4+y2-4y+4=2x2+2y2+162+4xy-36y-36xx2+y2+4xy-32y-32x+154=0
∴ Equation of the hyperbola = x2+y2+4xy-32y-32x+154=0

Page No 27.13:

Question 3:

Find the eccentricity, coordinates of the foci, equations of directrices and length of the latus-rectum of the hyperbola
(i) 9x2 − 16y2 = 144
(ii) 16x2 − 9y2 = −144
(iii) 4x2 − 3y2 = 36
(iv) 3x2y2 = 4
(v) 2x2 − 3y2 = 5.

Answer:

(i) Equation of the hyperbola:
9x2-16y2=144
This can be rewritten in the following way:
x216-y29=1
This is the standard equation of a hyperbola, where a2=16 and b2=9.

b2=a2(e2-1)9=16(e2-1)e2-1=916e2=2516e=54
Coordinates of the foci are given by ±ae,0, i.e. ±5,0.
Equation of directrices:
x=±ae
 x=±4545x±16=0
Length of the latus rectum of the hyperbola is 2b2a.
Length of the latus rectum = 2×94=92

(ii) Equation of the hyperbola:
16x2-9y2=-144
This can be rewritten in the following way:
x29-y216=-1
This is the standard equation of a hyperbola, where a2=9 and b2=16.

a2=b2(e2-1)9=16(e2-1)e2-1=916e2=2516e=54
Coordinates of foci are given by 0,±ae, i.e. 0,±5.
Equation of the directrices:
y=±ae
 y=±4545y±16=0
Length of the latus rectum of the hyperbola = 2a2b
Length of the latus rectum = 2×94=92

(iii) Equation of the hyperbola:
4x2-3y2=36
This can be rewritten in the following way:
4x236-3y236=1x29-y212=1
This is the standard equation of a hyperbola, where a2=9 and b2=12.

b2=a2(e2-1)12=9(e2-1)e2-1=43e2=73e=73
Coordinates of the foci are given by ±ae,0, i.e. ±21,0.
Equation of the directrices:
x=±ae
 x=±3737x±33=0
Length of the latus rectum of the hyperbola is 2b2a.
2×123=8

(iv) Equation of the hyperbola:
3x2-y2=4
This can be rewritten in the following way:
3x24-y24=1x243-y24=1
This is the standard equation of a hyperbola, where a2=43 and b2=4.

b2=a2(e2-1)4=43(e2-1)e2-1=3e2=4e=2
Coordinates of the foci are given by ±ae,0, i.e. ±433,0.
Equation of the directrices:
x=±ae
 x=±4323x±1=0
Length of the latus rectum of the hyperbola = 2b2a
2×443=43

(v) Equation of the hyperbola:
2x2-3y2=5
This can be rewritten in the following manner:
2x25-3y25=1x252-y253=1
This is the standard equation of a hyperbola, where a2=52 and b2=53.

b2=a2(e2-1)53=52(e2-1)e2-1=23e2=53e=53
Coordinates of the foci are given by ±ae,0, i.e. ±566,0.
Equation of the directrices:
x=±ae
 x=±5253x=±322x±3 = 0
Length of the latus rectum of the hyperbola is 2b2a.

2×5352=10325

Page No 27.13:

Question 4:

Find the axes, eccentricity, latus-rectum and the coordinates of the foci of the hyperbola 25x2 − 36y2 = 225.

Answer:

Equation of the hyperbola:
25x2-36y2=225
This equation can be rewritten in the following way:
25x2225-36y2225=1x29-y222536=1
This is the standard equation of the hyperbola, where a2=9 and b2=22536.
Length of the transverse = 2a=2×3=6
Length of the conjugate axis = 2b=2×156=5

Eccentricity of the hyperbola is calculated using b2=a2(e2-1).

22536=9e2-1e2-1=2536e2=6136e=616
Length of the latus rectum =2b2a=2×225363=256
The coordinates of the foci are given by ±ae,0.
±612,0

Page No 27.13:

Question 5:

Find the centre, eccentricity, foci and directrices of the hyperbola
(i) 16x2 − 9y2 + 32x + 36y − 164 = 0
(ii) x2y2 + 4x = 0
(iii) x2 − 3y2 − 2x = 8.

Answer:

(i) The equation 16x2-9y2+32x+36y-164=0 can be simplified in the following way:
16(x2+2x)-9y2-4y=16416(x2+2x+1)-9y2-4y+4=164+16-3616(x+1)2-9y-22=144(x+1)29-(y-2)216=1
Thus, the centre is -1,2.
Eccentricity of the hyperbola = a2+b2a=9+163=53
Foci = -1±5,2=-6,2,4,2

Equation of the directrices:
x+1=±aex=±3×35-1x=±95-15x-4=0 or 5x+14=0

(ii) The equation x2-y2+4x=0 can be simplified in the following manner:
(x2+4x+4)-y2=4x+22-y2=4(x+2)24-y21=1
Thus, the centre is -2,0.
Eccentricity of the hyperbola = a2+b2a=4+12=52
Foci = -2±5,0
Equation of the directrices:
x+2=±aex+2=±5

(iii) The equation x2-3y2-2x=8 can be simplified in the following manner:
(x2-2x+1)-3y2=8+1x-12-3y2=9(x-1)29-y23=1
Thus, the centre is 1,0.
Eccentricity of the hyperbola = a2+b2a=9+33=233
Foci = 1±23,0
Equation of the directrices:
x-1=±aex=1±233

Page No 27.13:

Question 6:

Find the equation of the hyperbola, referred to its principal axes as axes of coordinates, in the following cases:
(i) the distance between the foci = 16 and eccentricity = 2
(ii) conjugate axis is 5 and the distance between foci = 13
(iii) conjugate axis is 7 and passes through the point (3, −2).

Answer:

(i) The distance between the foci is 2ae.

2ae=16ae=8

 e=2
 a2=8a=42
 
Also, b2=a2(e2-1)b2=32(2-1)b2=32
Therefore, the standard form of the hyperbola is given below:
 x232-y232=1x2-y2=32

(ii) The distance between the foci is 2ae.
 
 2ae=13ae=132
Length of the conjugate axis, 2b=5
 b=52
 
Also, b2=a2(e2-1)522=1322-a2a2=169-254a2=1444=36a=6
Therefore, the standard form of the hyperbola is x236-4y225=1.
or  25x2-144y2=900


(iii) Length of the conjugate axis, 2b=7
b=72
Let the equation of the hyperbola be x2a2-y2b2=1.
It passes through 3,-2.

 32a2-(-2)2722=132a2-1649=19a2=1649+19a2=6549a2=44165

Therefore, the standard form of the hyperbola is 65x2441-4y249=1.
or 65x2-36y2=441



Page No 27.14:

Question 7:

Find the equation of the hyperbola whose
(i) foci are (6, 4) and (−4, 4) and eccentricity is 2.
(ii) vertices are (−8, −1) and (16, −1) and focus is (17, −1)
(iii) foci are (4, 2) and (8, 2) and eccentricity is 2.
(iv) vertices are at (0 ± 7) and foci at 0,±283.
(v) vertices are at (± 6, 0) and one of the directrices is x = 4. [NCERT EXEMPLAR]
(vi) foci at (± 2, 0) and eccentricity is 3/2.   [NCERT EXEMPLAR]

Answer:

(i) The centre of the hyperbola is the midpoint of the line joining  the two focii.
So, the coordinates of the centre are 6-42,4+42, i.e. 1, 4.
Let 2a and 2b be the length of the transverse and conjugate axis and let e be the eccentricity.

x-12a2-y-42b2=1
Distance between the two focii = 2ae


2ae=6+42+4-422ae=10ae=5a=52

Also, b2=ae2-a2b2=25-254b2=754

Equation of the hyperbola is given below:
4x-1225-4y-4275=14x2-2x+125-y2-8y+1675=14x2-8x+425-4y2-32y+6475=134x2-8x+4-4y2-32y+64=7512x2-24x+12-4y2+32y-64=7512x2-24x-4y2+32y-127=0

(ii) The centre of the hyperbola is given below:
-8+162,-1-12=4,-1
If the other focus is S'm,n, then it is calculated in the following way:
4=17+m2m=-9 And -1=-1+n2n=-1
Thus, the other focus is -9,-1.

Distance between the vertices: 2a=16+82+-1+122a=24a=12

Distance between the foci:
2c=17+92+-1+122c=26c=13

Also, c2=a2+b2b2=169-144b2=25

Equation of the hyperbola is given below:
x-42144-y+1225=1x2-8x+16144-y2+2y+125=125x2-200x+400-144y2+288y+144=360025x2-200x-144y2-288y-3344=0

(iii) The centre of the hyperbola is the midpoint of the line joining the two focii.
So, the coordinates of the centre are 4+82,2+22, i.e. 6, 2.
Let 2a and 2b be the length of the transverse and conjugate axis. Let e be the eccentricity.

x-62a2-y-22b2=1
Distance between the two focii = 2ae


2ae=4-82+2-222ae=4ae=2a=1

Also, b2=ae2-a2b2=4-1b2=3

Equation of the hyperbola:
x-621-y-223=1x2-12x+361-y2-4y+43=13x2-12x+36-y2-4y+4=33x2-36x+108-y2+4y-4=33x2-y2-36x+4y+101=0


(iv) The Vertices of the hyperbola are 0,±7.
∴ b=7
The foci is 0,±283.
∴ be=283


Also, a2=b2e2-1a2=2832-49a2=3439
Therefore, the equation of the hyperbola is-9x2343+y249=1.

(v) The Vertices of the hyperbola are ±6, 0.
∴ a=6
a2 = 36
Now, x = 4
ae=4e=32         a=6
Now,
 ae2=a2+b26×322=62+b281-36=b2b2=45


Therefore, the equation of the hyperbola isx236-y245=1.

(vi) The foci of the hyperbola are ±2, 0.
∴ ae=2a=2×23=43a2=169
Now,
 ae2=a2+b222=432+b24-169=b2b2=209


Therefore, the equation of the hyperbola is given by
9x216-9y220=1x24-y25=49

Page No 27.14:

Question 8:

Find the eccentricity of the hyperbola, the length of whose conjugate axis is 34 of the length of transverse axis.

Answer:

The lengths of the conjugate axis and the transverse axis are 2b and 2a, respectively.
We have:
2b=34×2ab=34a
Using the relation b2=a2(e2-1), we get:

342a2=a2(e2-1)916=e2-1e2=2516e=54
Therefore, the eccentricity of the hyperbola is 54.

Page No 27.14:

Question 9:

Find the equation of the hyperboala whose
(i) focus is at (5, 2), vertex at (4, 2) and centre at (3, 2)
(ii) focus is at (4, 2), centre at (6, 2) and e = 2.

Answer:

(i) The equation of the hyperbola with centre (x0,y0) is given by
x-x02a2-y-y02b2=1Focus = ae+x0, y0

 Vertex = (a+x0, y0)

 ae=2 and a=1b2=22-a2b2=22-12b2=3

x-321-y-223=13x-32-y-22=3


(ii) The equation of the hyperbola with centre (x0,y0) is given by
x-x02a2-y-y02b2=1Focus = ae+x0, y0

 ae=-2a=-1b2=22-a2b2=-22--12b2=3

x-621-y-223=13x-62-y-22=3

Page No 27.14:

Question 10:

If P is any point on the hyperbola whose axis are equal, prove that SP. S'P = CP2.

Answer:

Equation of the hyperbola:
x2a2-y2b2=1

If the axes of the hyperbola are equal, then a=b.
Then, equation of the hyperbola becomes x2-y2=a2.

   b2=a2e2-1a2=a2e2-11=e2-1e2=2e=2
Thus, the centre C0,0 and the focus are given by S2a,0 and S'-2a,0, respectively.
Let Pα,β be any point on the parabola.
So, it will satisfy the equation.
α2-β2=a2

 SP2=2a-α2+β2=2a2+α2-22aα+β2

S'P2=-2a-α2+β2=2a2+α2+22aα+β2

Now, SP2.S'P2=2a2+α2-22aα+β22a2+α2+22aα+β2=4a4+4a2α2+β2+α2+β22-8a2α2=4a2a2-2α2+4a2α2+β2+α2+β22=4a2α2-β2-2α2+4a2α2+β2+α2+β22=-4a2α2+β2+4a2α2+β2+α2+β22=α2+β22=CP4
∴ SP.S'P=CP2

Page No 27.14:

Question 11:

In each of the following find the equations of the hyperbola satisfying the given conditions:
(i) vertices (± 2, 0), foci (± 3, 0)
(ii) vertices (0, ± 5), foci (0, ± 8)
(iii) vertices (0, ± 3), foci (0, ± 5)
(iv) foci (± 5, 0), transverse axis = 8
(v) foci (0, ± 13), conjugate axis = 24
(vi) foci (± 35, 0), the latus-rectum = 8
(vii) foci (± 4, 0), the latus-rectum = 12
(viii) vertices (± 7, 0), e=43
(ix) foci (0, ± 10), passing through (2, 3)
(x) foci (0, ± 12), latus-rectum = 36

Answer:

(i) The vertices of the hyperbola are ±2,0 and the foci are ±3,0.
Thus, the value of a=2 and ae=3.
Now, using the relation b2=a2(e2-1), we get:
b2=9-4b2=5
Thus, the equation of the hyperbola is x24-y25=1.

(ii) The vertices of the hyperbola are 0,±5 and the foci are 0,±8
Thus, the value of a=5 and ae=8.
Now, using the relation b2=a2(e2-1), we get:
b2=64-25b2=39
Thus, the equation of the hyperbola is -x239+y225=1.

(iii) The vertices of the hyperbola are 0,±3 and the foci are 0,±5.
Thus, the value of a=3 and ae=5.
Now, using the relation b2=a2(e2-1), we get:
b2=25-9b2=16
Thus, the equation of the hyperbola is  -x216+y29=1.

(iv) The foci of  the hyperbola  are ±5,0 and the transverse axis is 8.
Thus, the value of  ae=5 and 2a = 8.
a=4
Now, using the relation b2=a2(e2-1), we get:
b2=25-16b2=9
Thus, the equation of the hyperbola isx216-y29=1

(v) The foci of the hyperbola are 0,±13 and the conjugate axis is 24.
Thus, the value of  ae=13 and 2b = 24.
b = 12

Now, using the relation b2=a2(e2-1), we get:
a2=169-144a2=25
Thus, the equation of the hyperbola is-x2144+y225=1

(vi) The foci of the hyperbola are ±35,0 and the latus rectum is 8.
Thus, the value of  ae=35 
and 2b2a=8b2=4a
Now, using the relation b2=a2(e2-1), we get:
4a=45-a2a2+4a-45=0a-5a+9=0a=-9, 5
 b2=-36 or 20
Since negative value is not possible, it is equal to 20.
Thus, the equation of the hyperbola isx225-y220=1

(vii) The foci of the hyperbola are ±4,0 and the latus rectum is 12.
Thus, the value of  ae=4
and 2b2a=12b2=6a
Now, using the relation b2=a2(e2-1), we get:
6a=16-a2a2+6a-16=0a-2a+8=0a=2, or -8
 b2=12 or -48
Since negative value is not possible, its value is 12.
Thus, the equation of the hyperbola isx24-y212=1

(viii) The vertices of hyperbola are ±7,0 and eccentricity is 43
Thus, the value of a=7.
Now, using the relation b2=a2(e2-1), we get:
b2=49169-1b2=49×79=3439
Thus, the equation of the hyperbola isx249-9y2343=1.

(ix) The foci of hyperbola are 0,±10 that pass through 2,3.

Thus, the value of ae=10.By squaring both the sides, we get:ae2=10a2+b2=10b2=10-a2 
Let the equation of the hyperbola be y2a2-x2b2=1.
It passes through 2,3.
32a2-2210-a2=190-9a2-4a2=10a2-a4a4-23a2+90=0a2-18a2-5=0a2=18,5
Now, b2=-8 or 5
If we neglect the negative value,  then b2 = 5.
Thus, the equation of the hyperbola isy25-x25=1.

(x) The foci of the hyperbola are 0,±12 and the latus rectum is 36.
Thus, the value of  ae=12.
and 2b2a=36b2=18a

Now, using the relation b2=a2(e2-1), we get:
18a=144-a2a2+18a-144=0a+24a-6=0a=-24, or 6
 b2=-432 or 108              (but negative value is not possible)

Thus, the equation of the hyperbola isy236-x2108=1.

Page No 27.14:

Question 12:

If the distance between the foci of a hyperbola is 16 and its ecentricity is 2, then obtain its equation.

Answer:

We have
2ae=16ae=8a=82=42a2=32
Now,
 ae2=a2+b282=32+b264-32=b2b2=32


Therefore, the equation of the hyperbola is given by
x232-y232=1x2-y2=32

Page No 27.14:

Question 13:

Show that the set of all points such that the difference of their distances from (4, 0) and (− 4,0) is always equal to 2 represents a hyperbola.

Answer:

Let the point be P(xy)
x-42+y-02-x+42+y-02=2x-42+y-022=2+x+42+y-022x-42+y2=4+x+42+y2+4x+42+y-02x-42-x+42=4+4x+42+y-02
-16x=4+4x+42+y-02-16x-4=4x+42+y-02-44x+1=4x+42+y-02-4x+1=x+42+y-0216x2+8x+1=x2+8x+16+y215x2-y2=15x21-y215=1
Which is the equation of a hyperbola.



Page No 27.18:

Question 1:

Write the eccentricity of the hyperbola 9x2 − 16y2 = 144.

Answer:

Equation of the hyperbola:
9 x2 - 16y 2 = 144                         ..... (1)

Equation (1) can be rewritten in the following way:
9x2144 - 16y 2144 = 144144x216 - y 29 = 1
x242-y232=1
This becomes the standard equation of the hyperbola with its major axis a=4 and minor axis b=3.
Eccentricity, e = a2+b2a
Substituting the value of a and b, we get:
e = 42+324    =16+94    =54
Therefore, the eccentricity is 54.

Page No 27.18:

Question 2:

Write the eccentricity of the hyperbola whose latus-rectum is half of its transverse axis.

Answer:

The lengths of the latus rectum and the transverse axis are 2b2a and 2a, respectively. ​
According to the given statement, length of the latus rectum is half of its trasverse axis.

2b2a = 12×2a2b2a = a   2b2=a2

Eccentricity, e = a2+b2a
Substituting the value b2=a22, we get:
e = a2+a22a    =a32a    =32
Therefore, the eccentricity is 32.

Page No 27.18:

Question 3:

Write the coordinates of the foci of the hyperbola 9x2 − 16y2 = 144.

Answer:

Equation of the hyperbola:
9x2-16y2=144 ..... (1)

This equation (1) can be rewritten in the following way:
9x2144-16y2144=1x216-y29=1


This becomes a standard form of the hyperbola with transverse axis a=4 and conjugate axis b=3.

Eccentricity of the hyperbola, e=a2+b2a                                                    =42+324                                                    =16+94                                                    =54
The foci of the hyperbola are of the form ae,0 and 
-ae,0.
Therefore, the foci are 5,0 and -5,0.4

Page No 27.18:

Question 4:

Write the equation of the hyperbola of eccentricity 2, if it is known that the distance between its foci is 16.

Answer:

The foci of the hyperbola are of the form ae,0 and -ae,0.

Distance between the foci = ae- (-ae2+02                                                =2ae2                                                =2ae
Distance between the foci is 16 and eccentricity of the hyperbola is 2.

  2ae=1622a=16a=42


Now, b2=a2(e2-1)  b2=422((2)2-1)  b2=32

Equation of the hyperbola is given below:
x2422-y232=1x232-y232=1

Page No 27.18:

Question 5:

If the foci of the ellipse x216+y2b2=1 and the hyperbola x2144-y281=125 coincide, write the value of b2.

Answer:

Equation of the hyperbola:
x2144-y281=125

Simplifying the above equation, we get:
25x2144-25y281=1x214425-y28125=1x21252-y2952=1
This is the standard form of hyperbola with a=125and b=95.

Eccentricity of the hyperbola, e=1252+952125                                      =1512
The foci of the hyperbola are of the form ae,0 and -ae,0.
So, foci of the hyperbola are 3,0 and -3,0.
The foci of the hyperbola coincide with the foci of the ellipse.

For ellipse,
a=4

Using the relationa2-b2=ae, we get:
42-b2=316-b2=9b2=7
Therefore, the value of b2 is 7.

Page No 27.18:

Question 6:

Write the length of the latus-rectum of the hyperbola 16x2 − 9y2 = 144.

Answer:

Equation of the hyperbola:
16x2-9y2=144
This equation can be rewritten in the following way:
16x2144-9y2144=1x29-y216=1x232-y242=1
This is the standard form of a hyperbola with a=3 and b=4.
Length of the latus rectum = 2b2a 
Substituting the value of and b, we get:
Length of the ​latus rectum  =2×423=323

Page No 27.18:

Question 7:

If the latus-rectum through one focus of a hyperbola subtends a right angle at the farther vertex, then write the eccentricity of the hyperbola.

Answer:

The standard equation of hyperbola is given below:
x2a2-y2b2=1

Let the latus rectum be QR, which passes through one of the focus. QR subtends a right angle at point P.

It passes through the focus ae,±y.
ae2a2-y2b2=1y2b2=ae2a2-1y2b2=e2-1y=±be2-1
Now, the slope of PQ is be2-1ae+a and the slope of PR is-be2-1ae+a.
The lines PQ and PR are perpendicular, so the product of the slope is -1.

be2-1ae+a×-be2-1ae+a=-1b2e2-1=ae+a2b2e2-1=a2e+12e2-12=e+12e4-2e2+1=e2+2e+1e4-3e2-2e=0ee3-3e-2=0ee-2e+1e+1=0

∴ e=0,-1,2

The value of e can neither be negative nor zero.

Therefore, the value of eccentricity is 2.

Page No 27.18:

Question 8:

Write the distance between the directrices of the hyperbola x = 8 sec θ, y = 8 tan θ.

Answer:

We have:
x=8secθ, y=8tanθ 

On squaring and subtracting, we get:x2-y2=64sec2θ-64tan2θx2-y2=64x264-y264=1

a = b = 8
Distance between the directrices of  hyperbola is 2a2a2+b2.

 2×6464+64=12882=162=82

Page No 27.18:

Question 9:

Write the equation of the hyperbola whose vertices are (± 3, 0) and foci at (± 5, 0).

Answer:

The vertices of the hyperbola and the foci are ±a,0 and ±ae,0, respectively.

 a=3 and ae=5 

Using the relation b2=a2e2-1, we get:

b2=ae2-a2b2=25-9b2=16b=4
Therefore, the equation of hyperbola is x29-y216=1.

Page No 27.18:

Question 10:

If e1 and e2 are respectively the eccentricities of the ellipse x218+y24=1 and the hyperbola x29-y24=1, then write the value of 2 e12 + e22.

Answer:

The standard form of the ellipse is x218+y24=1, where a2=18 and b2=4.
So, the eccentricity is calculated in the following way:
b2=a2(1-e12)4=18(1-e12)29=1-e12e12=79
The standard form of the hyperbola is x29-y24=1, where a2=9 and b2=4.
So, the eccentricity is calculated in the following way:
b2=a2(e22-1)4=9(e22-1)49=e22-1e22=139

 2e12+e22=2×79+139                  =279                  =3

Page No 27.18:

Question 1:

Equation of the hyperbola whose vertices are (± 3, 0) and foci at (± 5, 0), is
(a) 16x2 − 9y2 = 144
(b) 9x2 − 16y2 = 144
(c) 25x2 − 9y2 = 225
(d) 9x2 − 25y2 = 81

Answer:

(a) 16x2 − 9y2 = 144

The vertices of the hyperbola are ±3,0 and foci are ±5,0.
Thus, the values of a and ae are 3 and 5, respectively.

Now, using the relation b2=a2(e2-1), we get:
b2=25-9b2=16
Equation of the hyperbola is given below:
 x29-y216=116x2-9y2=144

Page No 27.18:

Question 2:

If e1 and e2 are respectively the eccentricities of the ellipse x218+y24=1 and the hyperbola x29-y24=1, then the relation between e1 and e2 is
(a) 3 e12 + e22 = 2
(b) e12 + 2 e22 = 3
(c) 2 e12 + e22 = 3
(d) e12 + 3 e22 = 2

Answer:

(c) 2 e12 + e22 = 3
The standard form of the ellipse is x218+y24=1, where a2=18 and b2=4.
So, the eccentricity is calculated in the following way:
b2=a2(1-e12)4=18(1-e12)29=1-e12e12=79
The standard form of the hyperbola is x29-y24=1, where a2=9 and b2=4.
So, the eccentricity is calculated in the following way:
b2=a2(e22-1)4=9(e22-1)49=e22-1e22=139

 2e12+e22=2×79+139                  =279                  =3



Page No 27.19:

Question 3:

The distance between the directrices of the hyperbola x = 8 sec θ, y = 8 tan θ, is
(a) 82
(b) 162
(c) 42
(d) 62

Answer:

(a) 82

We have:
x=8secθ, y=8tanθ 

On squaring and subtracting:x2-y2= 8sec2θ - 8tan2θx2- y2=8x28-y28=1

∴ a = b = 8
Distance between the directrices of the hyperbola = 2a2a2+b2
 Distance between the directrices=2×6464+64                                                 =12882                                                 =162                                                 =82

Page No 27.19:

Question 4:

The equation of the conic with focus at (1, 1) directrix along xy + 1 = 0 and eccentricity 2 is
(a) xy = 1
(b) 2xy + 4x − 4y − 1= 0
(c) x2y2 = 1
(d) 2xy − 4x + 4y + 1 = 0

Answer:

(d) 2xy − 4x + 4y + 1 = 0

Let P(x,y) be any point on the hyperbola.
Then, the distance of any point from the focus is eccentricity times the distance from the directrix.

 x-12+y+12=2x-y+12

Squaring both the sides, we get:
(x-1)2+(y+1)2=x-y+12x2-2x+1+y2+1+2y=x2+y2+1-2xy-2y+2x2xy-4x+4y+1=0

Page No 27.19:

Question 5:

The eccentricity of the conic 9x2 − 16y2 = 144 is
(a) 54

(b) 43

(c) 45

(d) 7

Answer:

(a) 54

Standard form of a hyperbola = x216-y29=1
Here, a2=16 and y2=9

The eccentricity is calculated in the following way:
b2=a2e2-19=16e2-1e2-1=916e2=2516e=54

Page No 27.19:

Question 6:

A point moves in a plane so that its distances PA and PB from two fixed points A and B in the plane satisfy the relation PAPB = k (k ≠ 0), then the locus of P is
(a) a hyperbola
(b) a branch of the hyperbola
(c) a parabola
(d) an ellipse

Answer:

(a) a hyperbola
Let P(x,y) be any point on the hyperbola x2a2-y2b2=1.By definition, we have:PA=ex-ae=ex-aand PB=ex+ae=ex+a PB-PA=ex+a-ex-a=2a=k

Page No 27.19:

Question 7:

The eccentricity of the hyperbola whose latus-rectum is half of its transverse axis, is
(a)12

(b) 23

(c) 32

(d) none of these.

Answer:

(c) 32
The lengths of the latus rectum and the transverse axis are 2b2a and 2a, respectively. ​
According to the given statement, length of the latus rectum is half of its transverse axis.
2b2a = 12×2a2b2a = a   2b2=a2

Eccentricity, e = a2+b2a
Substituting the value b2=a22 , we get:
e = a2+a22a    =a32a    =32
∴ Eccentricity is 32

Page No 27.19:

Question 8:

The eccentricity of the hyperbola x2 − 4y2 = 1 is

(a) 32

(b) 52

(c) 23

(d) 25

Answer:

(b) 52

The equation of the hyperbola is x2-4y2=1.
This can be rewritten in the following way:
x21-y214=1
This is the standard form of a hyperbola, where a2=1 and b2=14.
The value of eccentricity is calculated in the following way:
b2=a2(e2-1)14=(e2-1)e2=54e=52

Page No 27.19:

Question 9:

The difference of the focal distances of any point on the hyperbola is equal to
(a) length of the conjugate axis
(b) eccentricity
(c) length of the transverse axis
(d) Latus-rectum

Answer:

(c) length of the transverse axis

Let Px,ybe any point on the hyperbola, and S, S' be the focus with coordinates ±ae,0.

S'P-SP=2a
Thus, the difference of the focal distances of any point on the hyperbola is equal to the length of the transverse axis.

Page No 27.19:

Question 10:

The foci of the hyperbola 9x2 − 16y2 = 144 are
(a) (± 4, 0)
(b) (0, ± 4)
(c) (± 5, 0)
(d) (0, ± 5)

Answer:

(c) (± 5, 0)

The equation of the hyperbola is given below:
9x2-16y2=144
This equation can be rewritten in the following way:
9x2144-16y2144=1x216-y29=1
This is the standard equation of a hyperbola, where a2=16 and b2=9.

The eccentricity is calculated in the following way:
b2=a2(e2-1)9=16(e2-1)916=e2-1e=54
Foci = ±ae,0=±5,0

Page No 27.19:

Question 11:

The distance between the foci of a hyperbola is 16 and its eccentricity is 2, then equation of the hyperbola is
(a) x2 + y2 = 32
(b) x2y2 = 16
(c) x2 + y2 = 16
(d) x2y2 = 32

Answer:

(d) x2y2 = 32

The distance between the foci is 2ae.

 2ae=16ae=8

 e=2
 a2=8a=42


Also, b2=a2(e2-1)b2=32(2-1)b2=32
Standard form of the hyperbola is given below:
 x232-y232=1x2-y2=32

Page No 27.19:

Question 12:

If e1 is the eccentricity of the conic 9x2 + 4y2 = 36 and e2 is the eccentricity of the conic 9x2 − 4y2 = 36, then
(a) e12e22 = 2
(b) 2 < e22e12 < 3
(c) e22e12 = 2
(d) e22e12 > 3

Answer:

(b) 2 < e22e12 < 3
The conic ​9x2+4y2=36 can rewritten in the following way:
9x236+4y236=1x24+y29=1
This is the standard equation of an ellipse.
 b2=a21-e129=41-e12 e12=-54
The conic ​9x2-4y2=36 can rewritten in the following way:
9x236-4y236=1x24-y29=1
This is the standard equation of a hyperbola.
 b2=a2e22-19=4e22-1e22=134

 e22-e12=134+54=2.5

Page No 27.19:

Question 13:

If the eccentricity of the hyperbola x2y2 sec2α = 5 is 3 times the eccentricity of the ellipse x2 sec2 α + y2 = 25, then α =

(a) π6

(b) π4

(c) π3

(d) π2

Answer:

(b) π4
The hyperbola x2-y2sec2α=5 can be rewritten in the following way:
x25-y25cos2α=1

This is the standard form of a hyperbola, where a2=5 and b2=5cos2α.

b2=a2e12-15cos2α=5e12-1e12=cos2α+1              .....1

The ellipse x2sec2α+y2=25 can be rewritten in the following way:
x225cos2α+y225=1

This is the standard form of an ellipse, where a2=25 and b2=25cos2α.

b2=a21-e22e22=1-cos2αe22=sin2α          ......(2)

According to the question,

cos2α+1=3sin2α2=4sin2αsinα=12α=π4

Page No 27.19:

Question 14:

The equation of the hyperbola whose foci are (6, 4) and (−4, 4) and eccentricity 2, is

(a) (x-1)225/4-(y-4)275/4=1

(b) (x+1)225/4-(y+4)275/4=1

(c) (x-1)275/4-(y-4)225/4=1

(d) none of these

Answer:

(a) (x-1)225/4-(y-4)275/4=1

The centre of the hyperbola is the midpoint of the line joining the two foci.
So, the coordinates of the centre are 6-42,4+42, i.e. 1, 4.
Let 2a and 2b be the length of the transverse and the conjugate axes, respectively. Also, let e be the eccentricity.

x-12a2-y-42b2=1

Now, distance between the two foci = 2ae

2ae=6+42+4-422ae=10ae=5a=52

Also, b2=ae2-a2b2=25-254b2=754

Equation of the hyperbola is given below:
x-1225/4-y-4275/4=1

Page No 27.19:

Question 15:

The length of the straight line x − 3y = 1 intercepted by the hyperbola x2 − 4y2 = 1 is

(a) 65

(b) 325

(c) 625

(d) none of these

Answer:

(c) 625

The point of intersection of x-3y=1and the hyperbola x2-4y2=1 is calculated in the following way:
1+3y2-4y2=11+6y+9y2-4y2=15y2+6y=0y=0 or y=- 65
If y=0, then x=1.
If y=-65, then x=1+3×-65=-135.

So, the points are 1,0 and -135,-65.

∴ Length = 1+1352+0+652=625



Page No 27.20:

Question 16:

The latus-rectum of the hyperbola 16x2 − 9y2 = 144 is
(a) 16/3
(b) 32/3
(c) 8/3
(d) 4/3

Answer:

(b) 32/3

The standard form of the hyperbola 16x2-9y2=144 is x29-y216=1.
Here, a2=9, b2=16
Latus rectum of the hyperbola = 2b2a=2×163=323

Page No 27.20:

Question 17:

The foci of the hyperbola 2x2 − 3y2 = 5 are
(a) (±5/6,0)
(b) (± 5/6, 0)
(c) (±5/6,0)
(d) none of these

Answer:

(a) (±5/6,0)

The given equation of hyperbola is 2x2-3y2=5. It can be rewritten in the following way:
2x25-3y25=1x252-y253=1

This is the standard equation of a parabola, where a2=52 and b2=53.
The eccentricity can be calculated in the following way:
b2=a2e2-153=52e2-1e2-1=23e2=53e=53
Coordinates of the foci = ±ae,0=±56, 0

Page No 27.20:

Question 18:

The eccentricity the hyperbola x=a2t+1t, y=a2t-1t is
(a) 2
(b) 3
(c) 23
(d) 32

Answer:

(a) 2
We eliminate t in x=a2t+1t,y=a2t-1t.
On squaring both the sides in the equations x=a2t+1t,y=a2t-1t, we get:
4x2a2=t2+1t2+24x2a2-2=t2+1t2     ...1

Also, 4y2a2=t2+1t2-24y2a2+2=t2+1t2      ...2           

From (1) and (2), we get:
4x2a2-4y2a2=4x2a2-y2a2=1

This is the standard equation of a hyperbola, where a2=b2.

Eccentricity of the hyperbola, e=a2+a2a=2

Page No 27.20:

Question 19:

The equation of the hyperbola whose centre is (6, 2) one focus is (4, 2) and of eccentricity 2 is
(a) 3 (x − 6)2 − (y −2)2 = 3
(b) (x − 6)2 − 3 (y − 2)2 = 1
(c) (x − 6)2 − 2 (y −2)2 = 1
(d) 2 (x − 6)2 − (y − 2)2 = 1

Answer:

(a) 3 (x − 6)2 − (y −2)2 = 3

The equation of the hyperbola with centre (x0,y0) is given by
x-x02a2-y-y02b2=1Focus = ae+x0, y0


 ae=-2a=-1b2=22-a2b2=-22--12b2=3


 x-621-y-223=13x-62-y-22=3

Page No 27.20:

Question 20:

The locus of the point of intersection of the lines 3x-y-43λ=0 and 3λx+λy-43=0 is a hyperbola of eccentricity
(a) 1
(b) 2
(c) 3
(d) 4

Answer:

(b) 2

The equations of lines 3x-y-43λ=0 and 3λx+λy-43=0 can be rewritten as 3x-y=43λ and 3λx+λy=43 , respectively.

Multiplying the equations:

3λx2-λy2=48λ3λx248λ-λy248λ=1x216-y248=1
This is the standard equation of a hyperbola, where a2=16 and b2=48.

Eccentricity, e=a2+b2a2e=16+4816e=84e=2



View NCERT Solutions for all chapters of Class 14