NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.2 - Conic Section

Question 1: Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum.

$y^2 =12x$

Answer:

Given, a parabola with equation

$y^2 =12x$

This is parabola of the form $y^2=4ax$ which opens towards the right.

So,

By comparing the given parabola equation with the standard equation, we get,

$4a=12$

$a=3$

Hence,

Coordinates of the focus :

$(a,0)=(3,0)$

Axis of the parabola:

It can be seen that the axis of this parabola is X-Axis.

The equation of the directrix

$x=-a,\Rightarrow x=-3\Rightarrow x+3=0$

The length of the latus rectum:

$4a=4(3)=12$.

Question 2: Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum.

$x^2 = 6y$

Answer:

Given, a parabola with equation

$x^2 =6y$

This is parabola of the form $x^2=4ay$ which opens upward.

So,

By comparing the given parabola equation with the standard equation, we get,

$4a=6$

$a=\frac{3}{2}$

Hence,

Coordinates of the focus :

$(0,a)=\left (0,\frac{3}{2}\right)$

Axis of the parabola:

It can be seen that the axis of this parabola is Y-Axis.

The equation of the directrix

$y=-a,\Rightarrow y=-\frac{3}{2}\Rightarrow y+\frac{3}{2}=0$

The length of the latus rectum:

$4a=4(\frac{3}{2})=6$.

Question 3: Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum.

$y^2 = -8x$

Answer:

Given, a parabola with equation

$y^2 =-8x$

This is parabola of the form $y^2=-4ax$ which opens towards left.

So,

By comparing the given parabola equation with the standard equation, we get,

$-4a=-8$

$a=2$

Hence,

Coordinates of the focus :

$(-a,0)=(-2,0)$

Axis of the parabola:

It can be seen that the axis of this parabola is X-Axis.

The equation of the directrix

$x=a,\Rightarrow x=2\Rightarrow x-2=0$

The length of the latus rectum:

$4a=4(2)=8$.

Question 4: Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum.

$x^2 = -16y$

Answer:

Given, a parabola with equation

$x^2 =-16y$

This is parabola of the form $x^2=-4ay$ which opens downwards.

So,

By comparing the given parabola equation with the standard equation, we get,

$-4a=-16$

$a=4$

Hence,

Coordinates of the focus :

$(0,-a)=(0,-4)$

Axis of the parabola:

It can be seen that the axis of this parabola is Y-Axis.

The equation of the directrix

$y=a,\Rightarrow y=4\Rightarrow y-4=0$

The length of the latus rectum:

$4a=4(4)=16$.

Question 5: Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum.

$y^2 = 10x$

Answer:

Given, a parabola with equation

$y^2 =10x$

This is parabola of the form $y^2=4ax$ which opens towards the right.

So,

By comparing the given parabola equation with the standard equation, we get,

$4a=10$

$a=\frac{10}{4}=\frac{5}{2}$

Hence,

Coordinates of the focus :

$(a,0)=\left(\frac{5}{2},0\right)$

Axis of the parabola:

It can be seen that the axis of this parabola is X-Axis.

The equation of the directrix

$x=-a,\Rightarrow x=-\frac{5}{2}\Rightarrow x+\frac{5}{2}=0\Rightarrow 2x+5=0$

The length of the latus rectum:

$4a=4(\frac{5}{2})=10$.

Question 6: Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum.

$x^2 = -9y$

Answer:

Given, a parabola with equation

$x^2 =-9y$

This is parabola of the form $x^2=-4ay$ which opens downwards.

So

By comparing the given parabola equation with the standard equation, we get,

$-4a=-9$

$a=\frac{9}{4}$

Hence,

Coordinates of the focus :

$(0,-a)=\left (0,-\frac{9}{4}\right)$

Axis of the parabola:

It can be seen that the axis of this parabola is Y-Axis.

The equation of the directrix

$y=a,\Rightarrow y=\frac{9}{4}\Rightarrow y-\frac{9}{4}=0$

The length of the latus rectum:

$4a=4\left(\frac{9}{4}\right)=9$.

Question 7: Find the equation of the parabola that satisfies the given conditions:

Focus (6,0); directrix $x = - 6$

Answer:

Given, in a parabola,

Focus : (6,0) And Directrix : $x = - 6$

Here,

Focus is of the form (a, 0), which means it lies on the X-axis. And Directrix is of the form $x=-a$ which means it lies left to the Y-Axis.

These are the condition when the standard equation of a parabola is.$y^2=4ax$

Hence the Equation of Parabola is

$y^2=4ax$

Here, it can be seen that:

$a=6$

Hence the Equation of the Parabola is:

$\Rightarrow y^2=4ax\Rightarrow y^2=4(6)x$

$\Rightarrow y^2=24x$.

Question 8: Find the equation of the parabola that satisfies the given conditions:

Focus (0,–3); directrix $y = 3$

Answer:

Given,in a parabola,

Focus : Focus (0,–3); directrix $y = 3$

Here,

Focus is of the form (0,-a), which means it lies on the Y-axis. And Directrix is of the form $y=a$ which means it lies above X-Axis.

These are the conditions when the standard equation of a parabola is $x^2=-4ay$.

Hence the Equation of Parabola is

$x^2=-4ay$

Here, it can be seen that:

$a=3$

Hence the Equation of the Parabola is:

$\Rightarrow x^2=-4ay\Rightarrow x^2=-4(3)y$

$\Rightarrow x^2=-12y$.

Question 9: Find the equation of the parabola that satisfies the given conditions:

Vertex (0,0); focus (3,0)

Answer:

Given,

Vertex (0,0) And focus (3,0)

As vertex of the parabola is (0,0) and focus lies in the positive X-axis, The parabola will open towards the right, And the standard equation of such parabola is

$y^2=4ax$

Here it can be seen that $a=3$

So, the equation of a parabola is

$\Rightarrow y^2=4ax\Rightarrow y^2=4(3)x$

$\Rightarrow y^2=12x$.

Question 10: Find the equation of the parabola that satisfies the given conditions:

Vertex (0,0); focus (-2,0)

Answer:

Given,

Vertex (0,0) And focus (-2,0)

As vertex of the parabola is (0,0) and focus lies in the negative X-axis, The parabola will open towards left, And the standard equation of such parabola is

$y^2=-4ax$

Here it can be seen that $a=2$

So, the equation of a parabola is

$\Rightarrow y^2=-4ax\Rightarrow y^2=-4(2)x$

$\Rightarrow y^2=-8x$.

Question 11: Find the equation of the parabola that satisfies the given conditions:

Vertex (0,0) passing through (2,3) and axis is along x-axis.

Answer:

Given

The Vertex of the parabola is (0,0).

The parabola is passing through (2,3) and axis is along the x-axis, it will open towards right. and the standard equation of such parabola is

$y^2=4ax$

Now since it passes through (2,3)

$3^2=4a(2)$

$9=8a$

$a=\frac{8}{9}$

So the Equation of Parabola is ;

$\Rightarrow y^2=4\left(\frac{9}{8}\right)x$

$\Rightarrow y^2=\left(\frac{9}{2}\right)x$

$\Rightarrow 2y^2=9x$

Question 12: Find the equation of the parabola that satisfies the given conditions:

Vertex (0,0), passing through (5,2) and symmetric with respect to y-axis.

Answer:

Given a parabola,

with Vertex (0,0), passing through (5,2) and symmetric with respect to the y-axis.

Since the parabola is symmetric with respect to Y=axis, it's axis will ve Y-axis. and since it passes through the point (5,2), it must go through the first quadrant.

So the standard equation of such parabola is

$x^2=4ay$

Now since this parabola is passing through (5,2)

$5^2=4a(2)$

$25=8a$

$a=\frac{25}{8}$

Hence the equation of the parabola is

$\Rightarrow x^2=4\left ( \frac{25}{8} \right )y$

$\Rightarrow x^2=\left ( \frac{25}{2} \right )y$

$\Rightarrow 2x^2=25y$