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,

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

$\Rightarrow c=\sqrt{5}$

Hence, the Coordinates of the foci:

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

The vertices:

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

The length of the major axis:

$2a=2(3)=6$

The length of the minor axis:

$2b=2(2)=4$

The eccentricity :

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

The length of the latus rectum:

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

Question:10: Find the equation for the ellipse that satisfies the given conditions: Vertices (± 5, 0), foci (± 4, 0)

Answer:

Given, in an ellipse,

Vertices (± 5, 0), foci (± 4, 0)

Here Vertices and focus of the ellipse are on the X-axis, so the major axis of this ellipse will be the X-axis.

Therefore, the equation of the ellipse will be of the form:

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

Where $a$ and $b$ are the length of the semimajor axis and semiminor axis, respectively.

So, on comparing standard parameters( vertices and foci) with the given one, we get

$a=5$ and $c=4$

Now, as we know the relation,

$a^2=b^2+c^2$

$\Rightarrow b^2=a^2-c^2$

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

$\Rightarrow b=\sqrt{5^2-4^2}$

$\Rightarrow b=\sqrt{9}$

$\Rightarrow b=3$

Hence, The Equation of the ellipse will be :

$\frac{x^2}{5^2}+\frac{y^2}{3^2}=1$

$\Rightarrow \frac{x^2}{25}+\frac{y^2}{9}=1$ .

Question:11: Find the equation for the ellipse that satisfies the given conditions: Vertices (0, ± 13), foci (0, ± 5)

Answer:

Given, in an ellipse, Vertices (0, ± 13), foci (0, ± 5)

Here Vertices and focus of the ellipse are on the Y-axis, so the major axis of this ellipse will be the Y-axis.

Therefore, the equation of the ellipse will be of the form:

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

Where $a$ and $b$ are the length of the semimajor axis and semiminor axis, respectively.

So, on comparing standard parameters( vertices and foci) with the given one, we get

$a=13$ and $c=5$

Now, as we know the relation,

$a^2=b^2+c^2$

$\Rightarrow b^2=a^2-c^2$

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

$\Rightarrow b=\sqrt{{13}^2-5^2}$

$\Rightarrow b=\sqrt{169-25}$

$\Rightarrow b=\sqrt{144}$

$\Rightarrow b=12$

Hence, The Equation of the ellipse will be :

$\frac{x^2}{{12}^2}+\frac{y^2}{{13}^3}=1$

$\Rightarrow \frac{x^2}{144}+\frac{y^2}{169}=1$ .

Question:12: Find the equation for the ellipse that satisfies the given conditions: Vertices (± 6, 0), foci (± 4, 0)

Answer:

Given, in an ellipse, Vertices (± 6, 0), foci (± 4, 0)

Here Vertices and focus of the ellipse are on the X-axis, so the major axis of this ellipse will be the X-axis.

Therefore, the equation of the ellipse will be of the form:

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

Where $a$ and $b$ are the length of the semimajor axis and semiminor axis, respectively.

So, on comparing standard parameters( vertices and foci) with the given one, we get

$a=6$ and $c=4$

Now, as we know the relation,

$a^2=b^2+c^2$

$\Rightarrow b^2=a^2-c^2$

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

$\Rightarrow b=\sqrt{6^2-4^2}$

$\Rightarrow b=\sqrt{36-16}$

$\Rightarrow b=\sqrt{20}$

Hence, The Equation of the ellipse will be :

$\frac{x^2}{6^2}+\frac{y^2}{(\sqrt{20}{)}^2}=1$

$\Rightarrow \frac{x^2}{36}+\frac{y^2}{20}=1$ .

Question:13: Find the equation for the ellipse that satisfies the given conditions: Ends of major axis (± 3, 0), ends of minor axis (0, ± 2)

Answer:

Given, in an ellipse,

Ends of the major axis (± 3, 0), ends of minor axis (0, ± 2)

Here, the major axis of this ellipse will be the X-axis.

Therefore, the equation of the ellipse will be of the form:

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

Where $a$ and $b$ are the length of the semimajor axis and semiminor axis, respectively.

So, on comparing standard parameters( ends of the major and minor axis ) with the given one, we get $a=3$ and $b=2$

Hence, The Equation of the ellipse will be :

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

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

Question:14 Find the equation for the ellipse that satisfies the given conditions: Ends of major axis (0, ± $\sqrt{5}$ ), ends of minor axis (± 1, 0)

Answer:

Given, in an ellipse,

Ends of the major axis (0, ± $\sqrt{5}$ ), ends of minor axis (± 1, 0)

Here, the major axis of this ellipse will be the Y-axis.

Therefore, the equation of the ellipse will be of the form:

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

Where $a$ and $b$ are the length of the semimajor axis and semiminor axis, respectively.

So on comparing standard parameters( ends of the major and minor axis ) with the given one, we get $a=\sqrt{5}$ and $b=1$

Hence, The Equation of the ellipse will be :

$\frac{x^2}{1^2}+\frac{y^2}{(\sqrt{5}{)}^2}=1$

$\Rightarrow \frac{x^2}{1}+\frac{y^2}{5}=1$ .

Question:15 Find the equation for the ellipse that satisfies the given conditions: Length of major axis 26, foci (± 5, 0)

Answer:

Given, in an ellipse,

Length of major axis 26, foci (± 5, 0)

Here, the focus of the ellipse is on the X-axis, so the major axis of this ellipse will be the X-axis.

Therefore, the equation of the ellipse will be of the form:

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

Where $a$ and $b$ are the length of the semimajor axis and semiminor axis, respectively.

So on comparing standard parameters( Length of semimajor axis and foci) with the given one, we get $2a=26\Rightarrow a=13$ and $c=5$

Now, as we know the relation,

$a^2=b^2+c^2$

$\Rightarrow b^2=a^2-c^2$

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

$\Rightarrow b=\sqrt{{13}^2-5^2}$

$\Rightarrow b=\sqrt{144}$

$\Rightarrow b=12$

Hence, The Equation of the ellipse will be :

$\frac{x^2}{{13}^2}+\frac{y^2}{{12}^2}=1$

$\Rightarrow \frac{x^2}{169}+\frac{y^2}{144}=1$ .

Question:16: Find the equation for the ellipse that satisfies the given conditions: Length of minor axis 16, foci (0, ± 6).

Answer:

Given, in an ellipse,

Length of minor axis 16, foci (0, ± 6).

Here, the focus of the ellipse is on the Y-axis, so the major axis of this ellipse will be the Y-axis.

Therefore, the equation of the ellipse will be of the form:

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

Where $a$ and $b$ are the length of the semimajor axis and semiminor axis, respectively.

So, on comparing standard parameters( length of semi-minor axis and foci) with the given one, we get

$2b=16$
$\Rightarrow b=8$
and $c=6$

Now, as we know the relation,

$a^2=b^2+c^2$

$\Rightarrow a=\sqrt{b^2+c^2}$

$\Rightarrow a=\sqrt{8^2+6^2}$

$\Rightarrow a=\sqrt{64+36}$

$\Rightarrow a=\sqrt{100}$

$\Rightarrow a=10$

Hence, The Equation of the ellipse will be :

$\frac{x^2}{8^2}+\frac{y^2}{{10}^3}=1$

$\Rightarrow \frac{x^2}{64}+\frac{y^2}{100}=1$ .

Question:17 Find the equation for the ellipse that satisfies the given conditions: Foci (± 3, 0), a = 4

Answer:

Given, in an ellipse, V Foci (± 3, 0), a = 4

Here, foci of the ellipse is in the X-axis, so the major axis of this ellipse will be the X-axis.

Therefore, the equation of the ellipse will be of the form:

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

Where $a$ and $b$ are the length of the semimajor axis and semiminor axis, respectively.

So, on comparing standard parameters( vertices and foci) with the given one, we get

$a=4$ and $c=3$

Now, as we know the relation,

$a^2=b^2+c^2$

$\Rightarrow b^2=a^2-c^2$

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

$\Rightarrow b=\sqrt{4^2-3^2}$

$\Rightarrow b=\sqrt{7}$

Hence, The Equation of the ellipse will be :

$\frac{x^2}{4^2}+\frac{y^2}{(\sqrt{7}{)}^2}=1$

$\Rightarrow \frac{x^2}{16}+\frac{y^2}{7}=1$ .

Question:18 : Find the equation for the ellipse that satisfies the given conditions: b = 3, c = 4, centre at the origin; foci on the x-axis.

Answer:

Given: In an ellipse, b = 3, c = 4, centre at the origin; foci on the x-axis.

Here, foci of the ellipse is in the X-axis, so the major axis of this ellipse will be the X-axis.

Therefore, the equation of the ellipse will be of the form:

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

Where $a$ and $b$ are the length of the semimajor axis and semiminor axis, respectively.

Also Given,

$b=3$ and $c=4$

Now, as we know the relation,

$a^2=b^2+c^2$

$\Rightarrow a^2=3^2+4^2$

$\Rightarrow a^2=25$

$\Rightarrow a=5$

Hence, The Equation of the ellipse will be :

$\frac{x^2}{5^2}+\frac{y^2}{3^2}=1$

$\Rightarrow \frac{x^2}{25}+\frac{y^2}{9}=1$ .

Question:19: Find the equation for the ellipse that satisfies the given conditions: Centre at (0,0), major axis on the y-axis and passes through the points (3, 2) and (1,6).

Answer:

Given, in an ellipse

Centre at (0,0), major axis on the y-axis and passes through the points (3, 2) and (1,6).

Since the major axis of this ellipse is on the Y-axis, the equation of the ellipse will be of the form:

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

Where $a$ and $b$ are the length of the semimajor axis and semiminor axis, respectively.

Now, since the ellipse passes through points (3, 2)

$\frac{3^2}{b^2}+\frac{2^2}{a^2}=1$

$\Rightarrow 9a^2+4b^2=a^2{b}^2$

Since the ellipse also passes through points (1, 6).

$\frac{1^2}{b^2}+\frac{6^2}{a^2}=1$

$\Rightarrow a^2+36b^2=a^2{b}^2$

On solving these two equations, we get,

$a^2=40$ and $b^2=10$

Thus, the equation of the ellipse will be

$\frac{x^2}{10}+\frac{y^2}{40}=1$

Question:20: Find the equation for the ellipse that satisfies the given conditions: Major axis on the x-axis and passes through the points (4,3) and (6,2).

Answer:

Given, in an ellipse

Major axis is on the x-axis and passes through the points (4,3) and (6,2).

Since the major axis of this ellipse is on the X-axis, the equation of the ellipse will be of the form:

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

Where $a$ and $b$ are the length of the semimajor axis and semiminor axis, respectively.

Now, since the ellipse passes through the point (4,3)

$\frac{4^2}{a^2}+\frac{3^2}{b^2}=1$

$\Rightarrow 16b^2+9a^2=a^2{b}^2$

Since the ellipse also passes through the point (6, 2).

$\frac{6^2}{a^2}+\frac{2^2}{b^2}=1$

$\Rightarrow 4a^2+36b^2=a^2{b}^2$

On solving these two equations, we get,

$a^2=52$ and $b^2=13$

Thus, the equation of the ellipse will be

$\frac{x^2}{52}+\frac{y^2}{13}=1$