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#### Question 1:

Find the equation of the parabola whose:
(i) focus is (3, 0) and the directrix is 3x + 4y = 1
(ii) focus is (1, 1) and the directrix is x + y + 1 = 0
(iii) focus is (0, 0) and the directrix 2xy − 1 = 0
(iv) focus is (2, 3) and the directrix x − 4y + 3 = 0.

(i) Let P (x, y) be any point on the parabola whose focus is S (3, 0) and the directrix is 3x + 4y = 1.
Draw PM perpendicular to 3x + 4y = 1.
Then, we have:
$SP=PM\phantom{\rule{0ex}{0ex}}⇒S{P}^{2}=P{M}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-3\right)}^{2}+{\left(y-0\right)}^{2}={\left(\frac{3x+4y-1}{\sqrt{9+16}}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-3\right)}^{2}+{y}^{2}={\left(\frac{3x+4y-1}{5}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒25\left\{{\left(x-3\right)}^{2}+{y}^{2}\right\}={\left(3x+4y-1\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒\left(25{x}^{2}-150x+25{y}^{2}+225\right)=9{x}^{2}+16{y}^{2}+1+24xy-8y-6x\phantom{\rule{0ex}{0ex}}⇒16{x}^{2}+9{y}^{2}-24xy-144x+8y+224=0$

(ii) Let P (x, y) be any point on the parabola whose focus is S (1, 1) and the directrix is xy + 1 = 0.
Draw PM perpendicular to xy + 1 = 0.
Then, we have:
$SP=PM\phantom{\rule{0ex}{0ex}}⇒S{P}^{2}=P{M}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-1\right)}^{2}+{\left(y-1\right)}^{2}={\left|\frac{x+y+1}{\sqrt{1+1}}\right|}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-1\right)}^{2}+{\left(y-1\right)}^{2}={\left(\frac{x+y+1}{\sqrt{2}}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒2\left({x}^{2}+1-2x+{y}^{2}+1-2y\right)={x}^{2}+{y}^{2}+1+2xy+2y+2x\phantom{\rule{0ex}{0ex}}⇒\left(2{x}^{2}+2-4x+2{y}^{2}+2-4y\right)={x}^{2}+{y}^{2}+1+2xy+2y+2x\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}-2xy-6x-6y+3=0$

(iii) Let P (x, y) be any point on the parabola whose focus is S (0, 0) and the directrix is 2x − y − 1 = 0.
Draw PM perpendicular to 2x − y − 1 = 0.
Then, we have:
$SP=PM\phantom{\rule{0ex}{0ex}}⇒S{P}^{2}=P{M}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-0\right)}^{2}+{\left(y-0\right)}^{2}={\left|\frac{2x-y-1}{\sqrt{4+1}}\right|}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}={\left(\frac{2x-y-1}{\sqrt{5}}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒5{x}^{2}+5{y}^{2}=4{x}^{2}+{y}^{2}+1-4xy+2y-4x\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+4{y}^{2}+4xy-2y+4x-1=0$

(iv) Let P (x, y) be any point on the parabola whose focus is S (2, 3) and the directrix is x − 4y + 3 = 0.
Draw PM perpendicular to x − 4y + 3 = 0.
Then, we have:
$SP=PM\phantom{\rule{0ex}{0ex}}⇒S{P}^{2}=P{M}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-2\right)}^{2}+{\left(y-3\right)}^{2}={\left|\frac{x-4y+3}{\sqrt{1+16}}\right|}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-2\right)}^{2}+{\left(y-3\right)}^{2}={\left(\frac{x-4y+3}{\sqrt{17}}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒17\left({x}^{2}+4-4x+{y}^{2}-6y+9\right)={x}^{2}+16{y}^{2}+9-8xy-24y+6x\phantom{\rule{0ex}{0ex}}⇒\left(17{x}^{2}-68x-102y+17{y}^{2}+13×17\right)={x}^{2}+16{y}^{2}+9-8xy-24y+6x\phantom{\rule{0ex}{0ex}}⇒16{x}^{2}+{y}^{2}+8xy-74x-78y+212=0$

#### Question 2:

Find the equation of the parabola whose focus is the point (2, 3) and directrix is the line x − 4y + 3 = 0. Also, find the length of its latus-rectum.

Let P (x, y) be any point on the parabola whose focus is S (2, 3) and the directrix is x − 4y + 3 = 0.

Draw PM perpendicular to x − 4y + 3=0.
Then, we have:
$SP=PM\phantom{\rule{0ex}{0ex}}⇒S{P}^{2}=P{M}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-2\right)}^{2}+{\left(y-3\right)}^{2}={\left(\frac{x-4y+3}{\sqrt{1+16}}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-2\right)}^{2}+{\left(y-3\right)}^{2}={\left(\frac{x-4y+3}{\sqrt{17}}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒17\left({x}^{2}+4-4x+{y}^{2}-6y+9\right)={x}^{2}+16{y}^{2}+9-8xy-24y+6x\phantom{\rule{0ex}{0ex}}⇒\left(17{x}^{2}-68x+17{y}^{2}-102y+13×17\right)={x}^{2}+16{y}^{2}+9-8xy-24y+6x\phantom{\rule{0ex}{0ex}}⇒16{x}^{2}+{y}^{2}+8xy-74x-78y+212=0$

Length of the latus rectum = 2(Length of the perpendicular from the focus on the directrix)
= 2(Length of the perpendicular from (2, 3) on the directrix)
= $2\left|\frac{2-12+3}{\sqrt{16+1}}\right|=2\left|\frac{-7}{\sqrt{17}}\right|=2\left(\frac{7}{\sqrt{17}}\right)=\frac{14}{\sqrt{17}}$ units

#### Question 3:

Find the equation of the parabola if
(i) the focus is at (−6, −6) and the vertex is at (−2, 2)
(ii) the focus is at (0, −3) and the vertex is at (0, 0)
(iii) the focus is at (0, −3) and the vertex is at (−1, −3)
(iv) the focus is at (a, 0) and the vertex is at (a', 0)
(v) the focus is at (0, 0) and vertex is at the intersection of the lines x + y = 1 and xy = 3.

In a parabola, the vertex is the mid-point of the focus and the point of intersection of the axis and the directrix.

Let (x1, y1) be the coordinates of the point of intersection of the axis and directrix.

(i) It is given that the vertex and the focus of a parabola are (−2, 2) and (−6, −6), respectively.

∴ Slope of the axis of the parabola = $\frac{-6-2}{-6+2}=\frac{-8}{-4}=2$

Slope of the directrix = $\frac{-1}{2}$

Let the directrix intersect the axis at K (r, s).

∴ Required equation of the directrix:

$y-10=\frac{-1}{2}\left(x-2\right)$
$2y+x-22=0$.

Now, let P (x, y) be any point on the parabola whose focus is S (−6, −6), and the directrix is $2y+x-22=0$.

Draw PM perpendicular to $2x+y+22=0$.
Then, we have:
$SP=PM\phantom{\rule{0ex}{0ex}}⇒S{P}^{2}=P{M}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x+6\right)}^{2}+{\left(y+6\right)}^{2}={\left(\frac{2y+x-22}{\sqrt{5}}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒5\left({x}^{2}+12x+36+{y}^{2}+12y+36\right)=4{y}^{2}+{x}^{2}+484+4xy-88y-44x\phantom{\rule{0ex}{0ex}}⇒4{x}^{2}+{y}^{2}-4xy+104x+148y-124=0\phantom{\rule{0ex}{0ex}}⇒{\left(2x-y\right)}^{2}-4\left(26x+37y-31\right)=0$

(ii) It is given that the vertex and the focus of a parabola are (0, 0) and (0, −3), respectively.

Thus, the slope of the axis of the parabola cannot be defined.

Slope of the directrix = 0

Let the directrix intersect the axis at K (r, s).

∴ Required equation of directrix:
$y=3$

Let P (x, y) be any point on the parabola whose focus is S (0, −3) and the directrix is $y=3$.

Draw PM perpendicular to $y=3$.
Then, we have:
$SP=PM\phantom{\rule{0ex}{0ex}}⇒S{P}^{2}=P{M}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-0\right)}^{2}+{\left(y+3\right)}^{2}={\left(\frac{y-3}{\sqrt{1}}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}+6y+9={y}^{2}-6y+9\phantom{\rule{0ex}{0ex}}⇒{x}^{2}=-12y$

(iii) It is given that the vertex and the focus of a parabola are (−1, −3) and (0, −3), respectively.

Thus, the slope of the axis of the parabola is zero.

And, the slope of the directrix cannot be defined.

Let the directrix intersect the axis at K (r, s).

∴ Required equation of the directrix:
$x+2=0$

Let P (x, y) be any point on the parabola whose focus is S (0, −3) and the directrix is $x+2=0$.

Draw PM perpendicular to $x+2=0$.
Then, we have:
$SP=PM\phantom{\rule{0ex}{0ex}}⇒S{P}^{2}=P{M}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-0\right)}^{2}+{\left(y+3\right)}^{2}={\left|\frac{x+2}{\sqrt{1}}\right|}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}+6y+9={x}^{2}+4x+4\phantom{\rule{0ex}{0ex}}⇒{y}^{2}+6y-4x+5=0$

(iv) It is given that the vertex and the focus of a parabola are (a', 0) and (a, 0), respectively.

Thus, the slope of the axis of the parabola is zero.

And, the slope of the directrix cannot be defined.

Let the directrix intersect the axis at K (r, s).

∴ Required equation of the directrix is $x-2a\text{'}+a=0$.

Let P (x, y) be any point on the parabola whose focus is S (a, 0), and the directrix is $x-2a\text{'}+a=0$.

Draw PM perpendicular to $x-2a\text{'}+a=0$.
Then, we have:
$SP=PM\phantom{\rule{0ex}{0ex}}⇒S{P}^{2}=P{M}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-a\right)}^{2}+{\left(y-0\right)}^{2}={\left(\frac{x-2a\text{'}+a}{\sqrt{1}}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{y}^{2}={\left(x-2a\text{'}+a\right)}^{2}-{\left(x-a\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{y}^{2}={x}^{2}+4a{\text{'}}^{2}+{a}^{2}-4a\text{'}x-4aa\text{'}+2ax-{x}^{2}-{a}^{2}+2ax\phantom{\rule{0ex}{0ex}}⇒{y}^{2}=4a{\text{'}}^{2}-4a\text{'}x-4aa\text{'}+4ax\phantom{\rule{0ex}{0ex}}⇒{y}^{2}=-4\left(a\text{'}-a\right)\left(x-a\text{'}\right)$

(v) The point of intersection of is (2, −1).

Thus, the vertex and the focus of the parabola are (2, −1) and (0, 0), respectively.

∴ Slope of the axis of the parabola = $\frac{0+1}{0-2}=\frac{-1}{2}$

The slope of the directrix is 2.

Let the directrix intersect the axis at K (r, s).

The required equation of the directrix is $y+2=2\left(x-4\right)$, which can be rewritten as $y-2x+10=0$.

Let P (x, y) be any point on the parabola whose focus is S (0, 0) and directrix is $y-2x+10=0$.

Draw PM perpendicular to $y-2x+10=0$.
Then, we have:
$SP=PM\phantom{\rule{0ex}{0ex}}⇒S{P}^{2}=P{M}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-0\right)}^{2}+{\left(y-0\right)}^{2}={\left(\frac{y-2x+10}{\sqrt{5}}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒5{x}^{2}+5{y}^{2}={\left(y-2x+10\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+4{y}^{2}+4xy+40x-20y-100=0\phantom{\rule{0ex}{0ex}}⇒{\left(x+2y\right)}^{2}+40x-20y-100=0$

#### Question 4:

Find the vertex, focus, axis, directrix and latus-rectum of the following parabolas
(i) y2 = 8x
(ii) 4x2 + y = 0
(iii) y2 − 4y − 3x + 1 = 0
(iv) y2 − 4y + 4x = 0
(v) y2 + 4x + 4y − 3 = 0
(vi) y2 = 8x + 8y
(vii) 4 (y − 1)2 = − 7 (x − 3)
(viii) y2 = 5x − 4y − 9
(ix) x2 + y = 6x − 14

(i) Given:
y2 = 8x

On comparing the given equation with ${y}^{2}=4ax$:
$4a=8⇒a=2$

∴ Vertex = (0, 0)

Focus = (a, 0) = (2, 0)

Equation of the directrix:
x = −a
i.e. x = 2

Axis = y = 0

Length of the latus rectum = 4a = 8 units

(ii) Given:
4x2 + y = 0

$⇒\frac{-y}{4}={x}^{2}$

On comparing the given equation with ${x}^{2}=-4ay$:

$4a=\frac{1}{4}⇒a=\frac{1}{16}$

∴ Vertex = (0, 0)

Focus = (0, −a) = $\left(0,\frac{-1}{16}\right)$

Equation of the directrix:
ya
i.e. $y=\frac{1}{16}$

Axis = x = 0

Length of the latus rectum = 4a = $\frac{1}{4}$ units

(iii) Given:
y2 − 4y − 3x + 1 = 0

$⇒{\left(y-2\right)}^{2}-4-3x+1=0\phantom{\rule{0ex}{0ex}}⇒{\left(y-2\right)}^{2}=3\left(x+1\right)\phantom{\rule{0ex}{0ex}}⇒{\left(y-2\right)}^{2}=3\left(x-\left(-1\right)\right)$

Let $Y=y-2$, $X=x+1$
Then, we have:
${Y}^{2}=3X$

Comparing the given equation with ${Y}^{2}=4aX$:

$4a=3⇒a=\frac{3}{4}$

∴ Vertex = (X = 0, Y = 0) =

Focus = (X = a, Y = 0) =

Equation of the directrix:
X = −a
i.e$x+1=\frac{-3}{4}⇒x=\frac{-7}{4}$

Axis = Y = 0
i.e. $y-2=0⇒y=2$

Length of the latus rectum = 4a = 3 units

(iv) Given:
y2 − 4y + 4x = 0

$⇒{\left(y-2\right)}^{2}-4+4x=0\phantom{\rule{0ex}{0ex}}⇒{\left(y-2\right)}^{2}=-4\left(x-1\right)$

Let $Y=y-2$, $X=x-1$
Then, we have:
${Y}^{2}=-4X$

Comparing the given equation with ${Y}^{2}=-4aX$:

$4a=4⇒a=1$

∴ Vertex = (X = 0, Y = 0) =

Focus = (X = −a, Y = 0) =

Equation of the directrix:
X = a
i.e. $x-1=1⇒x=2$

Axis = Y = 0
i.e. $y-2=0⇒y=2$

Length of the latus rectum = 4a = 4 units

(v) Given:
y2 + 4y + 4x −3 = 0

$⇒{\left(y+2\right)}^{2}-4+4x-3=0\phantom{\rule{0ex}{0ex}}⇒{\left(y+2\right)}^{2}=-4\left(x-\frac{7}{4}\right)$

Let $Y=y+2$, $X=x-\frac{7}{4}$
Then, we have:
${Y}^{2}=-4X$

Comparing the given equation with ${Y}^{2}=-4aX$:
$4a=4⇒a=1$

∴ Vertex = (X = 0, Y = 0) =

Focus = (X = −a, Y = 0) =

Equation of the directrix:
X = a
i.e. $x-\frac{7}{4}=1⇒x=\frac{11}{4}$

Axis = Y = 0
i.e. $y+2=0⇒y=-2$

Length of the latus rectum = 4a = 4 units

(vi) Given:
y2 = 8x + 8y
$⇒{\left(y-4\right)}^{2}=8\left(x+2\right)$

Putting $Y=y-4$, $X=x+2$:
${Y}^{2}=8X$

On comparing the given equation with ${Y}^{2}=4aX$:
$4a=8⇒a=2$

∴ Vertex = (X = 0, Y = 0) =

Focus = (X = a, Y = 0) =
Equation of the directrix:
X = −a
i.e. $x+2=-2⇒x+4=0$
Axis = Y = 0
i.e. $y-4=0⇒y=4$
Length of the latus rectum = 4a = 8

(vii) Given:
4(y − 1)2 = − 7 (x − 3)
$⇒{\left(y-1\right)}^{2}=\frac{-7}{4}\left(x-3\right)$

Let $Y=y-1$, $X=x-3$
Then, we have:
${Y}^{2}=\frac{-7}{4}X$

Comparing the given equation with ${Y}^{2}=-4aX$:
$4a=\frac{7}{4}⇒a=\frac{7}{16}$

∴ Vertex = (X = 0, Y = 0) =

Focus = (X = −a, Y = 0) =

Equation of the directrix:
X = a
i.e. $x-3=\frac{7}{16}⇒x=\frac{55}{16}$

Axis = Y = 0
i.e.

Length of the latus rectum = 4a = $\frac{7}{4}$ units

(viii) Given:
y 2 = 5x − 4y − 9

$⇒{y}^{2}+4y=5x-9\phantom{\rule{0ex}{0ex}}⇒{\left(y+2\right)}^{2}=5x-5=5\left(x-1\right)$

Putting $Y=y+2$, $X=x-1$:
${Y}^{2}=5X$

Comparing the given equation with ${Y}^{2}=4aX$:
$4a=5⇒a=\frac{5}{4}$

∴ Vertex = (X = 0, Y = 0) =

Focus = (X = a, Y = 0) =

Equation of the directrix:
X = −a
i.e. $x-1=\frac{-5}{4}⇒x=\frac{-1}{4}$

Axis = Y = 0
i.e. $y+2=0⇒y=-2$

Length of the latus rectum = 4a = 5 units

(ix) Given:
x2 = 6xy−14
$⇒{\left(x-3\right)}^{2}=-y-14+9\phantom{\rule{0ex}{0ex}}⇒{\left(x-3\right)}^{2}=-y-5=-\left(y+5\right)$

Let $Y=y+5$, $X=x-3$
Then, we have:
${X}^{2}=-Y$

Comparing the given equation with ${X}^{2}=-4aY$:
$4a=1⇒a=\frac{1}{4}$

∴ Vertex = (X = 0, Y = 0) =

Focus = (X = 0, Y = −a) =

Equation of the directrix:
Y = a

i.e. $y+5=\frac{1}{4}⇒y=\frac{-19}{4}$

Axis = X = 0
i.e. $x-3=0⇒x=3$

Length of the latus rectum = 4a = 1 units

#### Question 5:

For the parabola y2 = 4px find the extremities of a double ordinate of length 8 p. Prove that the lines from the vertex to its extremities are at right angles.

The given equation of the parabola is y2 = 4px.

Let PQ be the double ordinate of length 8p of the parabola ${y}^{2}=4px$.

Then, we have:
PR = RQ = 4p

Let AR = x1
Then, the coordinates of P and Q are $\left({x}_{1},4p\right)$ and $\left({x}_{1},-4p\right)$, respectively.

Now, P lies on ${y}^{2}=4px\phantom{\rule{0ex}{0ex}}$.

${\left(4p\right)}^{2}=4p{x}_{1}$
$⇒{x}_{1}=4p$

So, the coordinates of P and Q are and $\left(4p,-4p\right)$, respectively.

The coordinates of A are (0, 0).

Now, ${m}_{1}{m}_{2}=-1$

Thus, AP is perpendicular to AQ.

Hence, the lines from the vertex to its extremities are at right angles.

#### Question 6:

Find the area of the triangle formed by the lines joining the vertex of the parabola ${x}^{2}=12y$ to the ends of its latus rectum.

The given equation of the parabola is x2 = 12y.

On comparing the given equation with ${x}^{2}=4ay$:
a = 3

Required area =

#### Question 7:

Find the coordinates of the point of intersection of the axis and the directrix of the parabola whose focus is (3, 3) and directrix is 3x − 4y = 2. Find also the length of the latus-rectum.

The given equation of the directrix is 3x − 4y = 2.                        (1)

∴ Slope of the directrix = $\frac{-3}{-4}=\frac{3}{4}$

Also, the axis is perpendicular to the directrix.
∴ Slope of the axis = $\frac{-4}{3}$

The focus lies on the axis of the parabola.
∴ Equation of the axis:
$\left(y-3\right)=\frac{-4}{3}\left(x-3\right)$
$\left(3y-9\right)=-4x+12$

$3y+4x-21=0$                         (2)

Solving equations (1) and (2):

Therefore, the intersection point of the axis and directrix is $\left(\frac{18}{5},\frac{11}{5}\right)$.
Also, length of the latus rectum = 2 (Length of the perpendicular from the focus on the directrix)
=$2\left|\frac{3\left(3\right)+\left(-4\right)3-2}{\sqrt{16+9}}\right|=2\left|\frac{-5}{\sqrt{16+9}}\right|=2$ square units

#### Question 8:

At what point of the parabola x2 = 9y is the abscissa three times that of ordinate?

Putting x = 3y in the given equation of the parabola:
$9{y}^{2}=9y\phantom{\rule{0ex}{0ex}}⇒9y\left(y-1\right)=0\phantom{\rule{0ex}{0ex}}⇒y=0,1$

At y = 0, x = 0
At y = 1, x = 3

Therefore, at (1, 3), the abscissa is three times that of the ordinate.

#### Question 9:

Find the equation of a parabola with vertex at the origin, the axis along x-axis and passing through (2, 3).

CASE I:
Let the equation of the required parabola be ${y}^{2}=4ax$                         (1)

Since (1) passes through (2, 3), we have:

$9=4a\left(2\right)⇒a=\frac{9}{8}$

Thus, the required equation is ${y}^{2}=\frac{4\left(9\right)x}{8}$, i.e. $2{y}^{2}=9x$.

CASE II:
Let the equation of the required parabola be ${y}^{2}=-4ax$                   (2)

Since (2) passes through (2, 3), we have:

$9=-4a\left(2\right)⇒a=\frac{-9}{8}$

Thus, the required equation is ${y}^{2}=\frac{-4\left(-9\right)x}{8}$, i.e. $2{y}^{2}=9x$.

Hence, in either case, the required equation of the parabola is $2{y}^{2}=9x$.

#### Question 10:

Find the equation of a parabola with vertex at the origin and the directrix, y = 2.

Let the equation of the directrix be y = a.

Comparing with y = 2:
a = 2

Equation of the parabola with directrix y =a is ${x}^{2}=-4ay$.

Hence, the required equation of the parabola is ${x}^{2}=-8y$.

#### Question 11:

Find the equation of the parabola whose focus is (5, 2) and having vertex at (3, 2).

Given:
The vertex and the focus of the parabola are (3, 2) and (5, 2), respectively.

∴ Slope of the axis of the parabola = 0

Slope of the directrix cannot be defined.

Let the directrix intersect the axis at K (r, s).

Required equation of the directrix is $x-1=0$, which can be rewritten as x = 1.

Let P (x, y) be any point on the parabola whose focus is S (5, 2) and the directrix is x =1.

Draw PM perpendicular to x  = 1.
Then, we have:
$SP=PM\phantom{\rule{0ex}{0ex}}⇒S{P}^{2}=P{M}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-5\right)}^{2}+{\left(y-2\right)}^{2}={\left(\frac{x-1}{1}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+25-10x+{y}^{2}+4-4y={x}^{2}+1-2x\phantom{\rule{0ex}{0ex}}⇒25-10x+{y}^{2}+4-4y-1+2x=0\phantom{\rule{0ex}{0ex}}⇒{y}^{2}-4y-8x+28=0$

#### Question 12:

The cable of a uniformly loaded suspension bridge hangs in the form of a parabola. The roadway which is horizontal and 100 m long is supported by vertical wires attached to the cable, the longest wire being 30 m and the shortest wire being 6 m. Find the length of a supporting wire attached to the roadway 18 m from the middle.

Let X'OX be the bridge and PAQ be the suspension cable.
The suspension cable forms a parabola with the vertex at (0, 6).

Let the equation of the parabola formed by the suspension cable be ${\left(x-0\right)}^{2}=4a\left(y-6\right)$.                                 (1)

It passes through P (−50, 30) and Q (50, 30).

$2500=4a\left(30-6\right)$
$4a=\frac{2500}{24}$

Putting the value of 4a in equation (1):

${x}^{2}=\frac{2500}{24}\left(y-6\right)$                                     (2)

Let LM be the supporting wire attached at M, which is 18 m from the mid-point (O) of the bridge.

Let the coordinates of L be (18, l).
It lies on the parabola (2).

${18}^{2}=\frac{2500}{24}\left(l-6\right)$

Hence, the length of the supporting wire attached to the roadway 18 m from the middle is 9.11 m.

#### Question 13:

Find the equations of the lines joining the vertex of the parabola y2 = 6x to the point on it which have abscissa 24. [NCERT EXEMPLAR]

Let A and B be points on the parabola y2 = 6x and OA, OB be the lines joining the vertex O to the points A and B whose abscissa are 24.

Now,
y2 = 6 × 24 = 144
y = ± 12
Therefore the coordinates of the points A and B are (24, 12) and (24, –12) respectively.
Hence the lines are given by
$y-0=±\frac{12-0}{24-0}\left(x-0\right)\phantom{\rule{0ex}{0ex}}⇒±2y=x\phantom{\rule{0ex}{0ex}}$

#### Question 14:

Find the coordinates of points on the parabola y2 = 8x whose focal distance is 4.  [NCERT EXEMPLAR]

We have y2 = 8x
⇒ y2 = 4(2)x
Comparing it with the general equation of parabola y2 = 4ax, we will get a = 2
Let the required point be (x1y1)
Now, Focal distance = 4
⇒ x1a = 4
⇒ x1 + 2 = 4
⇒ x1 = 2
Now, the point will satisfy the equation of parabola
∴ (y1)2 = 8(2) = 16
y1 = ± 4
Hence, the coordiantes of the points are (2, 4) and (2, −4).

#### Question 15:

Find the length of the line segment joining the vertex of the parabola y2 = 4ax and a point on the parabola where the line-segment makes an angle θ to the x-axis. [NCERT EXEMPLAR]

Let the coordinates of the point on the parabola be B (x1y1).
Let BO be the line segment
In right triangle AOB

∴ x1 = OBcosθ and y1 = OBsinθ
Now, the curve is passing through (x1y1)
∴ (y1)2 = 4a(x1)
⇒( OBsinθ)2 = 4a(OBcosθ)
$⇒{\mathrm{OB}}^{2}{\mathrm{sin}}^{2}\mathrm{\theta }=4a\mathrm{OBcos\theta }\phantom{\rule{0ex}{0ex}}⇒\mathrm{OB}=\frac{4a\mathrm{cos\theta }}{{\mathrm{sin}}^{2}\mathrm{\theta }}=4a\mathrm{cosec}\theta .\mathrm{cot}\theta$

#### Question 16:

If the points (0, 4) and (0, 2) are respectively the vertex and focus of a parabola, then find the equation of the parabola. [NCERT EXEMPLAR]

As the vertex and focus lie on y-axis, so y-axis is the axis of the parabola.
If the directrix meets the axis of the parabola at point Z, the AZ = AF = 2
OZ = OF + AZ + FA = 2 + 2 + 2 = 6
So, the equation of the directrix is y = 6
i.e., y − 6 = 0
Let P(xy) be any point in the plane of the focus and directrix and MP be the perpendicular
distance from P to the directrix, then P lies on parabola iff FP = MP

$⇒\sqrt{{\left(x-0\right)}^{2}+{\left(y-2\right)}^{2}}=\frac{\left|y-6\right|}{1}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}-4y+4={y}^{2}-12y+36\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+8y=32$

#### Question 17:

If the line y = mx + 1 is tangent to the parabola y2 = 4x, then find the value of m. [NCERT EXEMPLAR]

We have y2 = 4x
Substituting the value of  y = mx + 1 in y2 = 4x, we get
(mx + 1)2 = 4x
m2x2 + 2mx + 1 = 4x
m2x2 + (2m − 4)x + 1 = 0                       .....(1)
Since, a tangent touches the curve at a point, the roots of (1) must be equal.
D = 0
⇒ (2m − 4)2 − 4m2 = 0
⇒ 4m−16m + 16 − 4m2 = 0
m = 1

#### Question 1:

Write the axis of symmetry of the parabola y2 = x.

Clearly, the axis of symmetry of the given parabola is the x-axis.

#### Question 2:

Write the distance between the vertex and focus of the parabola y2 + 6y + 2x + 5 = 0.

Given:
${y}^{2}+6y+2x+5=0$

Let Y = y+3, $X=x-2$

From (1), we have:

${Y}^{2}=-2X$                            (2)

Putting $4a=2$:
$a=\frac{1}{2}$
Focus =
Vertex =

Thus, we have:
Focus = $\left(\frac{3}{2},-3\right)$
Vertex = $\left(2,-3\right)$

Distance between the vertex and the focus:

#### Question 3:

Write the equation of the directrix of the parabola x2 − 4x − 8y + 12 = 0.

Given:
x2 − 4x − 8y + 12 = 0

Let Y = y−1, $X=x-2$

∴ From (1), we have:

${X}^{2}=8Y$                                    (2)

Comparing with ${x}^{2}=4ay$:
$a=2$

Directrix = Y = −a

y − 1 = −a

⇒y = −a + 1
= −2 + 1
= −1

Therefore, the required equation of the directrix is $y=-1$.

#### Question 4:

Write the equation of the parabola with focus (0, 0) and directrix x + y − 4 = 0.

Let P (x, y) be any point on the parabola whose focus is S (0, 0) and the directrix is xy = 4.

Draw PM perpendicular to xy = 4.
Then, we have:
$SP=PM\phantom{\rule{0ex}{0ex}}⇒S{P}^{2}=P{M}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-0\right)}^{2}+{\left(y-0\right)}^{2}={\left(\frac{x+y-4}{\sqrt{1+1}}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}={\left(\frac{x+y-4}{\sqrt{2}}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}+2{y}^{2}={x}^{2}+{y}^{2}+16+2xy-8y-8x\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}-2xy+8x+8y-16=0$

#### Question 5:

Write the length of the chord of the parabola y2 = 4ax which passes through the vertex and is inclined to the axis at $\frac{\mathrm{\pi }}{4}$.

Let OP be the chord.

Let the coordinates of P be .

From the figure, we have:

$O{P}^{2}={{x}_{1}}^{2}+{{y}_{1}}^{2}$                             (1)

And, $\mathrm{tan}\frac{\mathrm{\pi }}{4}=\frac{{y}_{1}}{{x}_{1}}$

$⇒{x}_{1}={y}_{1}$                                   (2)

Also, lies on the parabola.

${{y}_{1}}^{2}=4a{x}_{1}$                              (3)

Using (2) and (3), we get:
${{x}_{1}}^{2}=4a{x}_{1}⇒{x}_{1}=4a$        ...(4)

∴ From (4), (1) and (2), we have:
$O{P}^{2}={\left(4a\right)}^{2}+{\left(4a\right)}^{2}=32{a}^{2}\phantom{\rule{0ex}{0ex}}⇒OP=4\sqrt{2}a$

Therefore, the length of the chord is .

#### Question 6:

If b and c are lengths of the segments of any focal chord of the parabola y2 = 4ax, then write the length of its latus-rectum.

Let S (a, 0) be the focus of the given parabola.

Let the end points of the focal chord be .

SP and SQ are segments of the focal chord with lengths b and c, respectively.

SP = b, SQ = c

Now, we have:
$\frac{1}{SP}+\frac{1}{SQ}=\frac{1}{a\left(1+{t}^{2}\right)}+\frac{{t}^{2}}{a\left(1+{t}^{2}\right)}=\frac{1}{a}$

$⇒\frac{1}{b}+\frac{1}{c}=\frac{1}{a}\phantom{\rule{0ex}{0ex}}⇒\frac{b+c}{bc}=\frac{1}{a}\phantom{\rule{0ex}{0ex}}⇒a=\frac{bc}{b+c}$

∴ Length of the latus rectum = 4a = $\frac{4bc}{b+c}$

#### Question 7:

PSQ is a focal chord of the parabola y2 = 8x. If SP = 6, then write SQ.

The coordinates of the focal chord are .
Comparing y2 = 8x with ${y}^{2}=4ax$:
a = 2

Therefore, the coordinates of the focus S is .

Given:
SP = 6

Thus, we have:

SQ = $\sqrt{{\left(2-\frac{2}{{t}^{2}}\right)}^{2}+\left(\frac{4}{{t}^{2}}\right)}$=$\sqrt{{\left(2-\frac{2}{2}\right)}^{2}+\left(\frac{4}{2}\right)}$ = 3

#### Question 8:

Write the coordinates of the vertex of the parabola whose focus is at (−2, 1) and directrix is the line x + y − 3 = 0.

Given:
The focus S is at (−2, 1) and the directrix is the line x + y − 3 = 0.

The slope of the line perpendicular to x + y − 3 = 0 is 1.

The axis of the parabola is perpendicular to the directrix and passes through the focus.

∴ Equation of the axis of the parabola = $y-1=1\left(x+2\right)$                 (1)

Intersection point of the directrix and axis is the intersection point of (1) and x + y − 3 = 0.

Let the intersection point be  K.

Therefore, the coordinates of K are (0, 3).

Let (h, k) be the coordinates of the vertex, which is the mid-point of the line segment joining K and the focus.

Hence, the coordinates of the vertex are (−1, 2).

#### Question 9:

If the coordinates of the vertex and focus of a parabola are (−1, 1) and (2, 3) respectively, then write the equation of its directrix.

Given:
The vertex and the focus of a parabola are (−1, 1) and (2, 3), respectively.

∴ Slope of the axis of the parabola = $\frac{3-1}{2+1}=\frac{2}{3}$

Slope of the directrix = $\frac{-3}{2}$

Let the directrix intersect the axis at K (r, s).

Now, required equation of the directrix:

$\left(y+1\right)=\frac{-3}{2}\left(x+4\right)$
$⇒3x+2y+14=0$

#### Question 10:

If the parabola y2 = 4ax passes through the point (3, 2), then find the length of its latus rectum.

We have y2 = 4ax
Since, the parabola is passing through the point (3, 2)
Hence, it will satisfy the equation of the parabola.
∴ 22 = 4(a)(3)
$⇒a=\frac{1}{3}$
Lenth of the latus ractum is given by
$4a\phantom{\rule{0ex}{0ex}}=4×\frac{1}{3}\phantom{\rule{0ex}{0ex}}=\frac{4}{3}$

#### Question 11:

Write the equation of the parabola whose vertex is at (−3,0) and the directrix is x + 5 = 0.

The general equation of the parabola is (yk)2 = 4a(xh)
Here, the (h, k) = (−3,0)
Now, the directrix is given by
x = h − a
⇒ −5 = −3 − a                            [âˆµ x + 5 = 0 ⇒ x = −5]
a = 2
Hence, the equation is given by
(y − 0)2 = 4(2)(x + 3)
y2 = 8 (x + 3)

#### Question 1:

The coordinates of the focus of the parabola y2x − 2y + 2 = 0 are
(a) (5/4, 1)
(b) (1/4, 0)
(c) (1, 1)
(d) none of these

(a)  (5/4, 1)

Given:
The equation of the parabola is y2x − 2y + 2 = 0.
$⇒{\left(y-1\right)}^{2}-1=\left(x-2\right)\phantom{\rule{0ex}{0ex}}⇒{\left(y-1\right)}^{2}=x-1$

Let

${Y}^{2}=X$

Comparing with ${Y}^{2}=4aX$:
$a=\frac{1}{4}$

Focus =

Hence, the focus is at (5/4, 1).

#### Question 2:

The vertex of the parabola (y + a)2 = 8a (xa) is
(a) (−a, −a)
(b) (a, −a)
(c) (−a, a)
(d) none of these

(b) (a, −a)

Given:
The equation of the parabola is (y + a)2 = 8a (xa).

Putting :

${Y}^{2}=8aX$

Vertex =

Hence, the vertex is at (a, −a).

#### Question 3:

If the focus of a parabola is (−2, 1) and the directrix has the equation x + y = 3, then its vertex is
(a) (0, 3)
(b) (−1, 1/2)
(c) (−1, 2)
(d) (2, −1)

(c)  (−1, 2)

Given:
The focus S is at (−2, 1) and the directrix is the line x + y − 3 = 0.

The slope of the line perpendicular to x + y − 3 = 0 is 1.

The axis of the parabola is perpendicular to the directrix and passes through the focus.

∴ Equation of the axis of the parabola = $y-1=1\left(x+2\right)$             (1)

Intersection point of the directrix and the axis is the intersection point of (1) and x + y − 3 = 0.

Let the intersection point be K.

Therefore, the coordinates of K will be (0, 3).

Let (h, k) be the coordinates of the vertex, which is the mid-point of the segment joining K and the focus.

Hence, the coordinates of the vertex are (−1, 2).

#### Question 4:

The equation of the parabola whose vertex is (a, 0) and the directrix has the equation x + y = 3a, is
(a) x2 + y2 + 2xy + 6ax + 10ay + 7a2 = 0
(b) x2 − 2xy + y2 + 6ax + 10ay − 7a2 = 0
(c) x2 − 2xy + y2 − 6ax + 10ay − 7a2 = 0
(d) none of these

(b) x2 − 2xy + y2 + 6ax + 10ay − 7a2 = 0

Given:
The vertex is at (a, 0) and the directrix is the line x + y = 3a.

The slope of the line perpendicular to x + y = 3a is 1.

The axis of the parabola is perpendicular to the directrix and passes through the vertex.

∴ Equation of the axis of the parabola = $y-0=1\left(x-a\right)$                           (1)

Intersection point of the directrix and the axis is the intersection point of (1) and x + y = 3a.

Let the intersection point be K.

Therefore, the coordinates of K are .

The vertex is the mid-point of the segment joining K and the focus (h, k).

Let P (x, y) be any point on the parabola whose focus is S (h, k) and the directrix is xy = 3a.

Draw PM perpendicular to xy = 3a.
Then, we have:
$SP=PM\phantom{\rule{0ex}{0ex}}⇒S{P}^{2}=P{M}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-0\right)}^{2}+{\left(y+a\right)}^{2}={\left(\frac{x+y-3a}{\sqrt{2}}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{\left(y+a\right)}^{2}={\left(\frac{x+y-3a}{\sqrt{2}}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}+2{y}^{2}+2{a}^{2}+4ay={x}^{2}+{y}^{2}+9{a}^{2}+2xy-6ax-6ay\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}-7{a}^{2}+10ay+6ax-2xy=0$

#### Question 5:

The parametric equations of a parabola are x = t2 + 1, y = 2t + 1. The cartesian equation of its directrix is
(a) x = 0
(b) x + 1 = 0
(c) y = 0
(d) none of these

(a) x = 0

Given:
x = t2 + 1         (1)
y = 2t + 1         (2)

From (1) and (2):
$x={\left(\frac{y-1}{2}\right)}^{2}+1$

On simplifying:
${\left(y-1\right)}^{2}=4\left(x-1\right)$

Let

${Y}^{2}=4X$

Comparing it with y2 = 4ax:
a = 1

Therefore, the equation of the directrix is X = −a , i.e. $x-1=-1⇒x=0$.

#### Question 6:

If the coordinates of the vertex and the focus of a parabola are (−1, 1) and (2, 3) respectively, then the equation of its directrix is
(a) 3x + 2y + 14 = 0
(b) 3x + 2y − 25 = 0
(c) 2x − 3y + 10 = 0
(d) none of these.

(a) 3x + 2y + 14 = 0

Given:
The vertex and the focus of a parabola are (−1, 1) and (2, 3), respectively.

∴ Slope of the axis of the parabola = $\frac{3-1}{2+1}=\frac{2}{3}$

Slope of the directrix = $\frac{-3}{2}$

Let the directrix intersect the axis at K (r, s).

Equation of the directrix:
$\left(y+1\right)=\frac{-3}{2}\left(x+4\right)$
$⇒3x+2y+14=0$

#### Question 7:

The locus of the points of trisection of the double ordinates of a parabola is a
(a) pair of lines
(b) circle
(c) parabola
(d) straight line

(c) parabola

Suppose PQ is a double ordinate of the parabola ${y}^{2}=4ax$.

Let R and S be the points of trisection of the double ordinates.

Let $\left(h,k\right)$ be the coordinates of R.

Then, we have:
OL = and RL = k

$\therefore RS=RL+LS=k+k=2k\phantom{\rule{0ex}{0ex}}⇒PR=RS=SQ=2k\phantom{\rule{0ex}{0ex}}⇒LP=LR+RP=k+2k=3k$

Thus, the coordinates of P are , which lie on ${y}^{2}=4ax$.

$9{k}^{2}=4ah$

Hence, the locus of the point (h, k) is $9{y}^{2}=4ax$,  i.e.  ${y}^{2}=\left(\frac{4a}{9}\right)x$, which represents a parabola.

#### Question 8:

The equation of the directrix of the parabola whose vertex and focus are (1, 4) and (2, 6) respectively is
(a) x + 2y = 4
(b) xy = 3
(c) 2x + y = 5
(d) x + 3y = 8

(a) x + 2y = 4

Given:
The vertex and the focus of a parabola are (1, 4) and (2, 6), respectively.

∴ Slope of the axis of the parabola = $\frac{6-4}{2-1}=2$

Slope of the directrix = $\frac{-1}{2}$

Let the directrix intersect the axis at K (r, s).

Equation of the directrix:

$\left(y-2\right)=\frac{-1}{2}\left(x-0\right)$
$⇒$x + 2y = 4

#### Question 9:

If V and S are respectively the vertex and focus of the parabola y2 + 6y + 2x + 5 = 0, then SV =
(a) 2
(b) 1/2
(c) 1
(d) none of these

(b) 1/2

Given:
The vertex and the focus of a parabola are V and S, respectively.

The given equation of parabola can be rewritten as follows:
${\left(y+3\right)}^{2}-9+5+2x=0$
$⇒{\left(y+3\right)}^{2}+2x=4\phantom{\rule{0ex}{0ex}}⇒{\left(y+3\right)}^{2}=4-2x\phantom{\rule{0ex}{0ex}}⇒{\left(y+3\right)}^{2}=-2\left(x-2\right)$

Let

Then, the equation of parabola becomes ${Y}^{2}=-2X$.

Vertex =

Comparing with y2 = 4ax:

$4a=2⇒a=\frac{1}{2}$

Focus =

SV = $\sqrt{{\left(2-\frac{3}{2}\right)}^{2}+{\left(-3+3\right)}^{2}}=\frac{1}{2}$

#### Question 10:

The directrix of the parabola x2 − 4x − 8y + 12 = 0 is
(a) y = 0
(b) x = 1
(c) y = − 1
(d) x = − 1

(c)  y = −1

Given:
x2 − 4x − 8y + 12 = 0
$⇒{\left(x-2\right)}^{2}-4-8y+12=0\phantom{\rule{0ex}{0ex}}⇒{\left(x-2\right)}^{2}=8y-8\phantom{\rule{0ex}{0ex}}⇒{\left(x-2\right)}^{2}=8\left(y-1\right)$

Putting X = x − 2, Y = y − 1:

${X}^{2}=8Y$

Comparing with ${X}^{2}=4aY$:
a = 2

Equation of the directrix:
$Y=-a$
$Y=-2\phantom{\rule{0ex}{0ex}}$
$⇒y-1=-2\phantom{\rule{0ex}{0ex}}⇒y=-2+1\phantom{\rule{0ex}{0ex}}⇒y=-1$

#### Question 11:

The equation of the parabola with focus (0, 0) and directrix x + y = 4 is
(a) x2 + y2 − 2xy + 8x + 8y − 16 = 0
(b) x2 + y2 − 2xy + 8x + 8y = 0
(c) x2 + y2 + 8x + 8y − 16 = 0
(d) x2y2 + 8x + 8y − 16 = 0

(a) x2 + y2 − 2xy + 8x + 8y − 16 = 0

Let P (x, y) be any point on the parabola whose focus is S (0, 0) and the directrix is xy = 4.

Draw PM perpendicular to xy = 4.
Then, we have:
$SP=PM\phantom{\rule{0ex}{0ex}}⇒S{P}^{2}=P{M}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-0\right)}^{2}+{\left(y-0\right)}^{2}={\left(\frac{x+y-4}{\sqrt{2}}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}={\left(\frac{x+y-4}{\sqrt{2}}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}+2{y}^{2}={x}^{2}+{y}^{2}+16+2xy-8x-8y\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}-2xy+8x+8y-16=0$

#### Question 12:

The line 2xy + 4 = 0 cuts the parabola y2 = 8x in P and Q. The mid-point of PQ is
(a) (1, 2)
(b) (1, −2)
(c) (−1, 2)
(d) (−1, −2)

(c) (−1, 2)

Let the coordinates of P and Q be and , respectively.

Slope of PQ = $\frac{2a{t}_{2}-2a{t}_{1}}{a{{t}_{2}}^{2}-a{{t}_{1}}^{2}}$               ......(1)

But, the slope of PQ is equal to the slope of 2xy + 4 = 0.

∴ Slope of PQ = $\frac{-2}{-1}=2$

From (1),

$\frac{2a{t}_{2}-2a{t}_{1}}{a{{t}_{2}}^{2}-a{{t}_{1}}^{2}}=2$                                .....(2)

Putting 4a = 8,
a = 2

∴ Focus of the given parabola = (a, 0) = .

Using equation (2):

$\frac{4\left({t}_{2}-{t}_{1}\right)}{2\left({{t}_{2}}^{2}-{{t}_{1}}^{2}\right)}=2$
$\frac{\left({t}_{2}-{t}_{1}\right)}{\left({{t}_{2}}^{2}-{{t}_{1}}^{2}\right)}=1$
$⇒{t}_{1}+{t}_{2}=1$
As, points P and Q lie on 2x-y+4=0

Let be the mid-point of PQ.

Then, we have:
${y}_{1}=\frac{2a{t}_{2}+2a{t}_{1}}{2}=2\left({t}_{1}+{t}_{2}\right)=2$
And, ${x}_{1}=\frac{a{{t}_{1}}^{2}+a{{t}_{2}}^{2}}{2}={{t}_{1}}^{2}+{{t}_{2}}^{2}=-1$

#### Question 13:

In the parabola y2 = 4ax, the length of the chord passing through the vertex and inclined to the axis at π/4 is
(a) $4\sqrt{2}a$
(b) $2\sqrt{2}a$
(c) $\sqrt{2}a$
(d) none of these

(a) $4\sqrt{2}a$

Let OP be the chord.

Let the coordinates of P be .

From the figure, we have:

$O{P}^{2}={{x}_{1}}^{2}+{{y}_{1}}^{2}$                   (1)
And, $\mathrm{tan}\frac{\mathrm{\pi }}{4}=\frac{{y}_{1}}{{x}_{1}}$
$⇒{x}_{1}={y}_{1}$                           (2)

Also, lies on the parabola.

${{y}_{1}}^{2}=4a{x}_{1}$                    (3)

Using (2) and (3):
${{x}_{1}}^{2}=4a{x}_{1}⇒{x}_{1}=4a$       (4)

∴ From (4), (1) and (2), we have:
$O{P}^{2}={\left(4a\right)}^{2}+{\left(4a\right)}^{2}=32{a}^{2}\phantom{\rule{0ex}{0ex}}⇒OP=4\sqrt{2}a$

Therefore, the length of the chord is .

#### Question 14:

The equation 16x2 + y2 + 8xy − 74x − 78y + 212 = 0 represents
(a) a circle
(b) a parabola
(c) an ellipse
(d) a hyperbola

(b) a parabola
Comparing the given equation with ax2 + by2 + 2hxy + 2gx+2fy + c =0, we get:

We have: ${h}^{2}=16=ab$

Thus, the given equation represents a parabola.

#### Question 15:

The length of the latus-rectum of the parabola y2 + 8x − 2y + 17 = 0 is
(a) 2
(b) 4
(c) 8
(d) 16

(c) 8

Given:
y2 + 8x − 2y + 17 = 0
$⇒{\left(y-1\right)}^{2}-1+8x+17=0\phantom{\rule{0ex}{0ex}}⇒{\left(y-1\right)}^{2}+8x+16=0\phantom{\rule{0ex}{0ex}}⇒{\left(y-1\right)}^{2}=-8\left(x+2\right)$

Let

${Y}^{2}=-8X$

Comparing with y2=4ax:
a = 2

Length of the latus rectum = 4a = 8 units

#### Question 16:

The vertex of the parabola x2 + 8x + 12y + 4 = 0 is
(a) (−4, 1)
(b) (4, −1)
(c) (−4, −1)
(d) (4, 1)

(a) (−4, 1)

Given:
x2 + 8x + 12y + 4 = 0
$⇒{\left(x+4\right)}^{2}-16+12y+4=0\phantom{\rule{0ex}{0ex}}⇒{\left(x+4\right)}^{2}+12y-12=0\phantom{\rule{0ex}{0ex}}⇒{\left(x+4\right)}^{2}=-12\left(y-1\right)$

Let

${X}^{2}=-12Y$

Vertex = $\left(X=0,Y=0\right)=\left(x+4=0,y-1=0\right)=\left(x=-4,y=1\right)$

Hence, the vertex is at (−4, 1).

#### Question 17:

The vertex of the parabola (y − 2)2 = 16 (x − 1) is
(a) (1, 2)
(b) (−1, 2)
(c) (1, −2)
(d) (2, 1)

(a) (1, 2)

Given:
(y − 2)2 = 16 (x − 1)

Let

${Y}^{2}=16X$

Vertex =

Hence, the vertex is at (1, 2).

#### Question 18:

The length of the latus-rectum of the parabola 4y2 + 2x − 20y + 17 = 0 is
(a) 3
(b) 6
(c) 1/2
(d) 9

(c) 1/2

Given:
4y2 + 2x − 20y + 17 = 0

$⇒{y}^{2}+\frac{x}{2}-5y+\frac{17}{4}=0\phantom{\rule{0ex}{0ex}}⇒{\left(y-\frac{5}{2}\right)}^{2}+\frac{x}{2}-2=0\phantom{\rule{0ex}{0ex}}⇒{\left(y-\frac{5}{2}\right)}^{2}=-1\left(\frac{x}{2}-2\right)\phantom{\rule{0ex}{0ex}}⇒{\left(y-\frac{5}{2}\right)}^{2}=\frac{-1}{2}\left(x-4\right)$

Let

∴ Length of the latus rectum = 4a = $\frac{1}{2}$ units

#### Question 19:

The length of the latus-rectum of the parabola x2− 4x − 8y + 12 = 0 is
(a) 4
(b) 6
(c) 8
(d) 10

(c) 8

Given:
x2− 4x − 8y + 12 = 0

$⇒{\left(x-2\right)}^{2}-8y+8=0\phantom{\rule{0ex}{0ex}}⇒{\left(x-2\right)}^{2}=8y-8=8\left(y-1\right)$

Let

${X}^{2}=8Y$

∴ Length of the latus rectum = 4a = 8 units

#### Question 20:

The focus of the parabola y = 2x2 + x is
(a) (0, 0)
(b) (1/2, 1/4)
(c) (−1/4, 0)
(d) (−1/4, 1/8)

(c) (−1/4, 0)

Given:
Equation of  the parabola = y = 2x2 + x
$⇒{x}^{2}+\frac{x}{2}=\frac{y}{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x+\frac{1}{4}\right)}^{2}=\frac{y}{2}+\frac{1}{16}\phantom{\rule{0ex}{0ex}}⇒{\left(x+\frac{1}{4}\right)}^{2}=\frac{8y+1}{16}\phantom{\rule{0ex}{0ex}}⇒{\left(x+\frac{1}{4}\right)}^{2}=\frac{1}{2}\left(y+\frac{1}{8}\right)$

Let
${X}^{2}=\frac{1}{2}Y$

Comparing with ${X}^{2}=4aY$:
$a=\frac{1}{8}$

Focus =

Hence, the focus is at (−1/4, 0).

#### Question 21:

Which of the following points lie on the parabola x2 = 4ay?
(a) x = at2, y = 2at
(b) x = 2at, y = at2
(c) x = 2at2, y = at
(d) x = 2at, y = at2

(d) x = 2at, y = at2

Substituting x = 2at, y = at2 in the given equation:
${\left(2at\right)}^{2}=4a\left(a{t}^{2}\right)\phantom{\rule{0ex}{0ex}}⇒4{a}^{2}{t}^{2}=4{a}^{2}{t}^{2}$

Hence, (2at, at2) lies on the parabola x2 = 4ay.

#### Question 22:

The equation of the parabola whose focus is (1, −1) and the directrix is x + y + 7 = 0 is
(a) x2 + y2 − 2xy − 18x − 10y = 0
(b) x2 − 18x − 10y − 45 = 0
(c) x2 + y2 − 18x − 10y − 45 = 0
(d) x2 + y2 − 2xy − 18x − 10y − 45 = 0

(d) x2 + y2 − 2xy − 18x − 10y − 45 = 0

Let P (x, y) be any point on the parabola whose focus is S (1, −1) and the directrix is xy + 7 = 0.

Draw PM perpendicular to xy + 7 = 0.
Then, we have:
$SP=PM\phantom{\rule{0ex}{0ex}}⇒S{P}^{2}=P{M}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-1\right)}^{2}+{\left(y+1\right)}^{2}={\left(\frac{x+y+7}{\sqrt{1+1}}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-1\right)}^{2}+{\left(y+1\right)}^{2}={\left(\frac{x+y+7}{\sqrt{2}}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒2\left({x}^{2}+1-2x+{y}^{2}+1+2y\right)={x}^{2}+{y}^{2}+49+2xy+14y+14x\phantom{\rule{0ex}{0ex}}⇒\left(2{x}^{2}+2-4x+2{y}^{2}+2+4y\right)={x}^{2}+{y}^{2}+49+2xy+14y+14x\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}-45-10y-2xy-18x=0$

Hence, the required equation is x2 + y2 − 2xy − 18x − 10y − 45 = 0.

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