NCERT Solutions for Class 11 Maths Chapter 10 Conic Sections

Question 1: Find the coordinates of the focus, the 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 a 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,

$4a=12$
$\Rightarrow a=3$

Hence,

Coordinates of the focus :

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

Axis of the parabola:

It can be seen that the axis of this parabola is the 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, the 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 a parabola of the form $x^2=4ay$, which opens upward.

So, by comparing the given parabola equation with the standard equation, we get,

$4a=6$

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

Hence,

Coordinates of the focus :

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

Axis of the parabola:

It can be seen that the axis of this parabola is the 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, the 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 a parabola of the form $y^2=-4ax$, which opens towards the left.

So, by comparing the given parabola equation with the standard equation, we get,

$-4a=-8$

$\Rightarrow a=2$

Hence,

Coordinates of the focus :

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

Axis of the parabola:

It can be seen that the axis of this parabola is the 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, the 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 a parabola of the form $x^2=-4ay$, which opens downward.

So, by comparing the given parabola equation with the standard equation, we get,

$-4a=-16$

$\Rightarrow a=4$

Hence,

Coordinates of the focus :

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

Axis of the parabola:

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

The equation of the directrix

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

The length of the latus rectum:

$4a=4\times 4=16$

Question 5: Find the coordinates of the focus, the 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 a 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,

$4a=10$

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

Hence,

Coordinates of the focus :

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

Axis of the parabola:

It can be seen that the axis of this parabola is the 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:

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

Question 6: Find the coordinates of the focus, the axis of the parabola, the equation of the directrix and the length of the latus rectum: $x^2=-9y$

Answer:

Given a parabola with an equation $x^2=-9y$

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

So, by comparing the given parabola equation with the standard equation, we get,

$-4a=-9$

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

Hence,

Coordinates of the focus :

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

Axis of the parabola:

It can be seen that the axis of this parabola is the 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.
Directrix is of the form $x=-a$, which means it lies left to the Y-Axis.

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

Hence, the Equation of the 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.
Directrix is of the form $y=a$, which means it lies above the X-axis.

These are the conditions when the standard equation of a parabola is $x^2=-4ay$. Hence, the Equation of the 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 the vertex of the parabola is (0,0) and the focus lies on the positive X-axis, the parabola will open towards the right. The standard equation of such a 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 the vertex of the parabola is (0,0) and the focus lies in the negative X-axis, the parabola will open towards the left, and the standard equation of such a 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 the x-axis.

Answer:

Given

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

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

$y^2=4ax$

Now, since it passes through (2,3)

$3^2=4a(2)$

$\Rightarrow 9=8a$

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

So the Equation of the 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 the 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 the Y-axis, its axis will be the Y-axis. and since it passes through the point (5,2), it must go through the first quadrant.

So the standard equation of such a parabola is

$x^2=4ay$

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

$5^2=4a(2)$

$\Rightarrow 25=8a$

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

Hence, the equation of the parabola is:

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

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

$\Rightarrow 2x^2=25y$

Question 1: Find the coordinates of the foci, the vertices, the length of the major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse: $\frac{x^2}{36}+\frac{y^2}{16}=1$

Answer:

The equation of the ellipse $\frac{x^2}{36}+\frac{y^2}{16}=1$

As we can see from the equation, the major axis is along the X-axis, and the minor axis is along the Y-axis.

On comparing the given equation with the standard equation of an ellipse, which is

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

We get,

$a=6$ and $b=4$

So,

$c=\sqrt{a^2-b^2}=\sqrt{6^2-4^2}=\sqrt{20}=2\sqrt{5}$

Hence,

Coordinates of the foci:

$(c,0)$ and $(-c,0)=(2\sqrt{5},0)$ and $(-2\sqrt{5},0)$

The vertices:

$(a,0)$ and $(-a,0)=(6,0)$ and $(-6,0)$

The length of the major axis:

$2a=2(6)=12$

The length of the minor axis:

$2b=2(4)=8$

The eccentricity :

$e=\frac{c}{a}=\frac{2\sqrt{5}}{6}=\frac{\sqrt{5}}{3}$

The length of the latus rectum:

$\frac{2b^2}{a}=\frac{2(4{)}^2}{6}=\frac{32}{6}=\frac{16}{3}$

Question 2: Find the coordinates of the foci, the vertices, the length of the major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse: $\frac{x^2}{4}+\frac{y^2}{25}=1$

Answer:

The equation of the ellipse $\frac{x^2}{4}+\frac{y^2}{25}=1$

As we can see from the equation, the major axis is along the Y-axis, and the minor axis is along the X-axis.

On comparing the given equation with the standard equation of such an ellipse, which is

$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$

We get,

$a=5$ and $b=2$

So,

$c=\sqrt{a^2-b^2}=\sqrt{5^2-2^2}=\sqrt{21}$

Hence,

Coordinates of the foci:

$(0,c)$ and $(0,-c)=(0,\sqrt{21})$ and $(0,-\sqrt{21})$

The vertices:

$(0,a)$ and $(0,-a)=(0,5)$ and $(0,-5)$

The length of the major axis:

$2a=2(5)=10$

The length of the minor axis:

$2b=2(2)=4$

The eccentricity :

$e=\frac{c}{a}=\frac{\sqrt{21}}{6}$

The length of the latus rectum:

$\frac{2b^2}{a}=\frac{2(2{)}^2}{5}=\frac{8}{5}$

Question 3: Find the coordinates of the foci, the vertices, the length of the major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse: $\frac{x^2}{16}+\frac{y^2}{9}=1$

Answer:

The equation of the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$

As we can see from the equation, the major axis is along the X-axis, and the minor axis is along the Y-axis.

On comparing the given equation with the standard equation of an ellipse, which is

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

We get,

$a=4$ and $b=3$

So,

$c=\sqrt{a^2-b^2}=\sqrt{4^2-3^2}=\sqrt{7}$

Hence, the Coordinates of the foci:

$(c,0)$ and $(-c,0)=(\sqrt{7},0)$ and $(-\sqrt{7},0)$

The vertices:

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

The length of the major axis:

$2a=2(4)=8$

The length of the minor axis:

$2b=2(3)=6$

The eccentricity :

$e=\frac{c}{a}=\frac{\sqrt{7}}{4}$

The length of the latus rectum:

$\frac{2b^2}{a}=\frac{2(3{)}^2}{4}=\frac{18}{4}=\frac{9}{2}$

Question 4: Find the coordinates of the foci, the vertices, the length of the major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse: $\frac{x^2}{25}+\frac{y^2}{100}=1$

Answer:

The equation of the ellipse $\frac{x^2}{25}+\frac{y^2}{100}=1$

As we can see from the equation, the major axis is along the Y-axis, and the minor axis is along the X-axis.

On comparing the given equation with the standard equation of such an ellipse, which is

$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$

We get,

$a=10$ and $b=5$

So,

$c=\sqrt{a^2-b^2}=\sqrt{{10}^2-5^2}=\sqrt{75}=5\sqrt{3}$

Hence, the Coordinates of the foci:

$(0,c)$ and $(0,-c)=(0,5\sqrt{3})$ and $(0,-5\sqrt{3})$

The vertices:

$(0,a)$ and $(0,-a)=(0,10)$ and $(0,-10)$

The length of the major axis:

$2a=2(10)=20$

The length of the minor axis:

$2b=2(5)=10$

The eccentricity :

$e=\frac{c}{a}=\frac{5\sqrt{3}}{10}=\frac{\sqrt{3}}{2}$

The length of the latus rectum:

$\frac{2b^2}{a}=\frac{2(5{)}^2}{10}=\frac{50}{10}=5$

Question 5: Find the coordinates of the foci, the vertices, the length of the major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse: $\frac{x^2}{49}+\frac{y^2}{36}=1$

Answer:

The equation of ellipse $\frac{x^2}{49}+\frac{y^2}{36}=1\Rightarrow \frac{x^2}{7^2}+\frac{y^2}{6^2}=1$

As we can see from the equation, the major axis is along the X-axis, and the minor axis is along the Y-axis.

On comparing the given equation with the standard equation of an ellipse, which is

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

We get,

$a=7$ and $b=6$ .

So,

$c=\sqrt{a^2-b^2}=\sqrt{7^2-6^2}=\sqrt{13}$

Hence, the Coordinates of the foci:

$(c,0)and(-c,0)=(\sqrt{13},0)and(-\sqrt{13},0)$

The vertices:

$(a,0)and(-a,0)=(7,0)and(-7,0)$

The length of the major axis:

$2a=2(7)=14$

The length of the minor axis:

$2b=2(6)=12$

The eccentricity :

$e=\frac{c}{a}=\frac{\sqrt{13}}{7}$

The length of the latus rectum:

$\frac{2b^2}{a}=\frac{2(6{)}^2}{7}=\frac{72}{7}$

Question 6: Find the coordinates of the foci, the vertices, the length of the major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse: $\frac{x^2}{100}+\frac{y^2}{400}=1$

Answer:

The equation of the ellipse $\frac{x^2}{100}+\frac{y^2}{400}=1\Rightarrow \frac{x^2}{{10}^2}+\frac{y^2}{{20}^2}=1$

As we can see from the equation, the major axis is along the Y-axis, and the minor axis is along the X-axis.

On comparing the given equation with the standard equation of such an ellipse, which is

$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$

We get,

$a=20$ and $b=10$ .

So,

$c=\sqrt{a^2-b^2}=\sqrt{{20}^2-{10}^2}$

$\Rightarrow c=\sqrt{300}=10\sqrt{3}$

Hence,

Coordinates of the foci:

$(0,c)and(0,-c)=(0,10\sqrt{3})and(0,-10\sqrt{3})$

The vertices:

$(0,a)and(0,-a)=(0,20)and(0,-20)$

The length of the major axis:

$2a=2(20)=40$

The length of the minor axis:

$2b=2(10)=20$

The eccentricity :

$e=\frac{c}{a}=\frac{10\sqrt{3}}{20}=\frac{\sqrt{3}}{2}$

The length of the latus rectum:

$\frac{2b^2}{a}=\frac{2(10{)}^2}{20}=\frac{200}{20}=10$

Question 7: Find the coordinates of the foci, the vertices, the length of the major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse: $36x^2+4y^2=144$

Answer:

The equation of the ellipse $36x^2+4y^2=144$

$\Rightarrow \frac{36}{144}{x}^2+\frac{4}{144}{y}^2=1$

$\Rightarrow \frac{1}{4}{x}^2+\frac{1}{36}{y}^2=1$

$\frac{x^2}{2^2}+\frac{y^2}{6^2}=1$

As we can see from the equation, the major axis is along the Y-axis, and the minor axis is along the X-axis.

On comparing the given equation with the standard equation of such an ellipse, which is

$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$

We get,

$a=6$ and $b=2$ .

So,

$c=\sqrt{a^2-b^2}=\sqrt{6^2-2^2}$

$\Rightarrow c=\sqrt{32}=4\sqrt{2}$

Hence, the Coordinates of the foci:

$(0,c)and(0,-c)=(0,4\sqrt{2})and(0,-4\sqrt{2})$

The vertices:

$(0,a)and(0,-a)=(0,6)and(0,-6)$

The length of the major axis:

$2a=2(6)=12$

The length of the minor axis:

$2b=2(2)=4$

The eccentricity :

$e=\frac{c}{a}=\frac{4\sqrt{2}}{6}=\frac{2\sqrt{2}}{3}$

The length of the latus rectum:

$\frac{2b^2}{a}=\frac{2(2{)}^2}{6}=\frac{8}{6}=\frac{4}{3}$

Question 8: Find the coordinates of the foci, the vertices, the length of the major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse: $16x^2+y^2=16$

Answer:

The equation of the ellipse $16x^2+y^2=16$

$\frac{16x^2}{16}+\frac{y^2}{16}=1$

$\frac{x^2}{1^2}+\frac{y^2}{4^2}=1$

As we can see from the equation, the major axis is along the Y-axis, and the minor axis is along the X-axis.

On comparing the given equation with the standard equation of such an ellipse, which is

$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$

We get,

$a=4$ and $b=1$ .

So,

$c=\sqrt{a^2-b^2}=\sqrt{4^2-1^2}$

$\Rightarrow c=\sqrt{15}$

Hence, the Coordinates of the foci:

$(0,c)and(0,-c)=(0,\sqrt{15})and(0,-\sqrt{15})$

The vertices:

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

The length of the major axis:

$2a=2(4)=8$

The length of the minor axis:

$2b=2(1)=2$

The eccentricity :

$e=\frac{c}{a}=\frac{\sqrt{15}}{4}$

The length of the latus rectum:

$\frac{2b^2}{a}=\frac{2(1{)}^2}{4}=\frac{2}{4}=\frac{1}{2}$

Question 9: Find the coordinates of the foci, the vertices, the length of the major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse: $4x^2+9y^2=36$

Answer:

The equation of the ellipse $4x^2+9y^2=36$

$\Rightarrow \frac{4x^2}{36}+\frac{9y^2}{36}=1$

$\Rightarrow \frac{x^2}{9}+\frac{y^2}{4}=1$

$\Rightarrow \frac{x^2}{3^2}+\frac{y^2}{2^2}=1$

As we can see from the equation, the major axis is along the X-axis, and the minor axis is along the Y-axis.

On comparing the given equation with the standard equation of an ellipse, which is

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

We get,

$a=3$ and $b=2$ .

So,

$c=\sqrt{a^2-b^2}=\sqrt{3^2-2^2}$