NCERT Solutions for Class 11 Maths Chapter 11 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,

$4a=12$
$⇒a=3$

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, 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,

$4a=6$

$⇒a=\frac{3}{2}$

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, 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,

$-4a=-8$

$⇒a=2$

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, 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 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:

$4a=4\times4=16$

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,

$4a=10$

$⇒a=\frac{10}{4}=\frac{5}{2}$

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:

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

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 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.
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 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 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, And 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)

$3^2=4a(2)$

$⇒9=8a$

$⇒a=\frac{9}{8}$

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 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 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)

$5^2=4a(2)$

$⇒25=8a$

$⇒a=\frac{25}{8}$

Hence the equation of the parabola is:

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

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

$\Rightarrow 2x^2=25y$

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,

$c=\sqrt{a^2-b^2}=\sqrt{6^2-4^2}=\sqrt{20}=2\sqrt5$

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,

$c=\sqrt{a^2-b^2}=\sqrt{4^2-3^2}=\sqrt7$

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,

$c=\sqrt{a^2-b^2}=\sqrt{10^2-5^2}=\sqrt{75}=5\sqrt3$

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,

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

$⇒c=\sqrt{300}=10\sqrt{3}$

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,

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

$⇒c=\sqrt{32}=4\sqrt{2}$

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,

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

$⇒c=\sqrt{15}$

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 $4x^2 + 9y^2 =36$

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

$⇒\frac{x^2}{9} + \frac{y^2}{4} = 1$

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

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,

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

$⇒c=\sqrt{5}$

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 in X-axis so the major axis of this ellipse will be 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,

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

$⇒b^2=a^2-c^2$

$⇒b=\sqrt{a^2-c^2}$

$⇒b=\sqrt{5^2-4^2}$

$⇒b=\sqrt{9}$

$⇒b=3$

Hence, The Equation of the ellipse will be :

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

$⇒\frac{x^2}{25}+\frac{y^2}{9}=1$ .

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 in Y-axis so the major axis of this ellipse will be 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,

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

$⇒b^2=a^2-c^2$

$⇒b=\sqrt{a^2-c^2}$

$⇒b=\sqrt{13^2-5^2}$

$⇒b=\sqrt{169-25}$

$⇒b=\sqrt{144}$

$⇒b=12$

Hence, The Equation of the ellipse will be :

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

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

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 in X-axis so the major axis of this ellipse will be 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,

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

$⇒b^2=a^2-c^2$

$⇒b=\sqrt{a^2-c^2}$

$⇒b=\sqrt{6^2-4^2}$

$⇒b=\sqrt{36-16}$

$⇒b=\sqrt{20}$

Hence, The Equation of the ellipse will be :

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

$⇒\frac{x^2}{36}+\frac{y^2}{20}=1$ .

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 :

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

$⇒\frac{x^2}{9}+\frac{y^2}{4}=1$ .

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 :

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

$⇒\frac{x^2}{1}+\frac{y^2}{5}=1$ .

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 in X-axis so the major axis of this ellipse will be 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,

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

$⇒b^2=a^2-c^2$

$⇒b=\sqrt{a^2-c^2}$

$⇒b=\sqrt{13^2-5^2}$

$⇒b=\sqrt{144}$

$⇒b=12$

Hence, The Equation of the ellipse will be :

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

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

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 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,

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

$⇒a=\sqrt{b^2+c^2}$

$⇒a=\sqrt{8^2+6^2}$

$⇒a=\sqrt{64+36}$

$⇒a=\sqrt{100}$

$⇒a=10$

Hence, The Equation of the ellipse will be :

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

$⇒\frac{x^2}{64}+\frac{y^2}{100}=1$ .

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 are in X-axis so the major axis of this ellipse will be 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,

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

$⇒b^2=a^2-c^2$

$⇒b=\sqrt{a^2-c^2}$

$⇒b=\sqrt{4^2-3^2}$

$⇒b=\sqrt{7}$

Hence, The Equation of the ellipse will be :

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

$⇒\frac{x^2}{16}+\frac{y^2}{7}=1$ .

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 are in X-axis so the major axis of this ellipse will be 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,

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

$⇒a^2=3^2+4^2$

$⇒a^2=25$

$⇒a=5$

Hence, The Equation of the ellipse will be :

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

$⇒\frac{x^2}{25}+\frac{y^2}{9}=1$ .

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)

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

$⇒{9a^2+4b^2}={a^2b^2}$

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

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

$⇒a^2+36b^2=a^2b^2$

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)

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

$⇒{16b^2+9a^2}={a^2b^2}$

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

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

$⇒4a^2+36b^2=a^2b^2$

On solving these two equations, we get,

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

Thus, The equation of the ellipse will be

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

Question:1 Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas: $\frac{x^2}{16} - \frac{y^2}{9} = 1$

Answer:
Given a Hyperbola equation, $\frac{x^2}{16} - \frac{y^2}{9} = 1$

It can also be written as $\frac{x^2}{4^2} - \frac{y^2}{3^2} = 1$

Comparing this equation with the standard equation of the hyperbola, we get,

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

We get,

$a=4$ and $b=3$

Now, As we know the relation in a hyperbola,

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

$⇒c=\sqrt{a^2+b^2}$

$⇒c=\sqrt{4^2+3^2}$

$⇒c=5$

Here as we can see from the equation the axis of the hyperbola is the X-axis.
So, Coordinates of the foci:

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

The Coordinates of vertices:

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

The Eccentricity:

$e=\frac{c}{a}=\frac{5}{4}$

The Length of the latus rectum :

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

Question:2 Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas: $\frac{y^2}{9} - \frac{x^2}{27} = 1$

Answer:

Given a Hyperbola equation, $\frac{y^2}{9} - \frac{x^2}{27} = 1$

It can also be written as $\frac{y^2}{3^2} - \frac{x^2}{(\sqrt{27})^2} = 1$

Comparing this equation with the standard equation of the hyperbola:

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

We get,

$a=3$ and $b=\sqrt{27}$

Now, As we know the relation in a hyperbola,

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

$⇒c=\sqrt{a^2+b^2}$

$⇒c=\sqrt{3^2+(\sqrt{27})^2}$

$⇒c=\sqrt{36}$

$⇒c=6$

Here as we can see from the equation the axis of the hyperbola is the Y-axis.
So, Coordinates of the foci:

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

The Coordinates of vertices:

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

The Eccentricity:

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

The Length of the latus rectum :

$\frac{2b^2}{a}=\frac{2(27)}{3}=\frac{54}{3}=18$

Question:3 Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas: $9 y^2 - 4 x^2 =36$

Answer:

Given a Hyperbola equation, $9 y^2 - 4 x^2 =36$

It can also be written as

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

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

Comparing this equation with the standard equation of the hyperbola:

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

We get,

$a=2$ and $b=3$

Now, As we know the relation in a hyperbola,

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

$⇒c=\sqrt{a^2+b^2}$

$⇒c=\sqrt{2^2+3^2}$

$⇒c=\sqrt{13}$

Hence, the Coordinates of the foci:

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

The Coordinates of vertices:

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

The Eccentricity:

$e=\frac{c}{a}=\frac{\sqrt{13}}{2}$

The Length of the latus rectum :

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

Question:4 Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas: $16x^2 - 9y^2 = 576$

Answer:

Given a Hyperbola equation, $16x^2 - 9y^2 = 576$

It can also be written as:

$\frac{16x^2}{576} - \frac{9y^2}{576} = 1$

$⇒\frac{x^2}{36} - \frac{y^2}{64} = 1$

$⇒\frac{x^2}{6^2} - \frac{y^2}{8^2} = 1$

Comparing this equation with the standard equation of the hyperbola:

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

We get,

$a=6$ and $b=8$

Now, As we know the relation in a hyperbola,

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

$⇒c=\sqrt{a^2+b^2}$

$⇒c=\sqrt{6^2+8^2}$

$⇒c=10$

Therefore, Coordinates of the foci:

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

The Coordinates of vertices:

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

The Eccentricity:

$e=\frac{c}{a}=\frac{10}{6}=\frac{5}{3}$

The Length of the latus rectum :

$\frac{2b^2}{a}=\frac{2(8)^2}{6}=\frac{128}{6}=\frac{64}{3}$

Question:5 Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas: $5y^2 - 9x^2 = 36$

Answer:

Given a Hyperbola equation, $5y^2 - 9x^2 = 36$

It can also be written as:

$\frac{5y^2}{36} - \frac{9x^2}{36} = 1$

$⇒\frac{y^2}{\frac{36}{5}} - \frac{x^2}{4} = 1$

$⇒\frac{y^2}{(\frac{6}{\sqrt{5}})^2} - \frac{x^2}{2^2} = 1$

Comparing this equation with the standard equation of the hyperbola:

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

We get,

$a=\frac{6}{\sqrt{5}}$ and $b=2$

Now, As we know the relation in a hyperbola,

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

$⇒c=\sqrt{a^2+b^2}$

$⇒c=\sqrt{(\frac{6}{\sqrt{5}})^2+2^2}$

$⇒c=\sqrt{\frac{56}{5}}$

$⇒c=2\sqrt{\frac{14}{5}}$

Here as we can see from the equation the axis of the hyperbola is the Y-axis.
So, Coordinates of the foci:

$(0,c)$ and $(0,-c)=\left(0,2\sqrt{\frac{14}{5}}\right)$ and $\left(0,-2\sqrt{\frac{14}{5}}\right)$

The Coordinates of vertices:

$(0,a) $ and $(0,-a)=\left(0,\frac{6}{\sqrt{5}}\right)$ and $\left(0,-\frac{6}{\sqrt{5}}\right)$

The Eccentricity:

$e=\frac{c}{a}=\frac{2\sqrt{\frac{14}{5}}}{\frac{6}{\sqrt{5}}}=\frac{\sqrt{14}}{3}$

The Length of the latus rectum :

$\frac{2b^2}{a}=\frac{2(4)}{\frac{6}{\sqrt{5}}}=\frac{4\sqrt{5}}{3}$

Question:6 Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas: $49y^2 - 16x^2 = 784$

Answer:

Given a Hyperbola equation, $49y^2 - 16x^2 = 784$

It can also be written as:

$\frac{49y^2}{784} - \frac{16x^2}{784} = 1$

$⇒\frac{y^2}{16} - \frac{x^2}{49} = 1$

$⇒\frac{y^2}{4^2} - \frac{x^2}{7^2} = 1$

Comparing this equation with the standard equation of the hyperbola:

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

We get,

$a=4$ and $b=7$

Now, As we know the relation in a hyperbola,

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

$⇒c=\sqrt{a^2+b^2}$

$⇒c=\sqrt{4^2+7^2}$

$⇒c=\sqrt{65}$

Therefore,

Coordinates of the foci:

$(0,c)$ and $(0,-c)=(0,\sqrt{65})$ and $(0,-\sqrt{65})$

The Coordinates of vertices:

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

The Eccentricity:

$e=\frac{c}{a}=\frac{\sqrt{65}}{4}$

The Length of the latus rectum :

$\frac{2b^2}{a}=\frac{2(49)}{4}=\frac{98}{4}=\frac{49}{2}$

Question:7 Find the equations of the hyperbola satisfying the given conditions: Vertices (± 2, 0), foci (± 3, 0)

Answer:

Given, in a hyperbola: Vertices (± 2, 0), foci (± 3, 0)

Here, Vertices and foci are on the X-axis so, the standard equation of the Hyperbola will be ;

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

By comparing the standard parameter (Vertices and foci) with the given one, we get

$a=2$ and $c=3$

Now, As we know the relation in a hyperbola

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

$⇒b^2=c^2-a^2$

$⇒b^2=3^2-2^2$

$⇒b^2=9-4=5$

Hence,The Equation of the hyperbola is $\frac{x^2}{4}-\frac{y^2}{5}=1$.

Question:8 Find the equations of the hyperbola satisfying the given conditions: Vertices (0, ± 5), foci (0, ± 8)

Answer:

Given, in a hyperbola: Vertices (0, ± 5), foci (0, ± 8)

Here, Vertices and foci are on the Y-axis so, the standard equation of the Hyperbola will be:

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

By comparing the standard parameter (Vertices and foci) with the given one, we get

$a=5$ and $c=8$

Now, As we know the relation in a hyperbola

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

$⇒b^2=c^2-a^2$

$⇒b^2=8^2-5^2$

$⇒b^2=64-25=39$

Hence, The Equation of the hyperbola is: $\frac{y^2}{25}-\frac{x^2}{39}=1$.

Question:9 Find the equations of the hyperbola satisfying the given conditions: Vertices (0, ± 3), foci (0, ± 5)

Answer:

Given, in a hyperbola Vertices (0, ± 3), foci (0, ± 5)

Here, Vertices and foci are on the Y-axis so, the standard equation of the Hyperbola will be:

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

By comparing the standard parameter (Vertices and foci) with the given one, we get

$a=3$ and $c=5$

Now, As we know the relation in a hyperbola

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

$⇒b^2=c^2-a^2$

$⇒b^2=5^2-3^2$

$⇒b^2=25-9=16$

Hence, The Equation of the hyperbola is $\frac{y^2}{9}-\frac{x^2}{16}=1$.

Question:10 Find the equations of the hyperbola satisfying the given conditions: Foci (± 5, 0), the transverse axis is of length 8.

Answer:

Given, in a hyperbola: Foci (± 5, 0), the transverse axis is of length 8.

Here, foci are on the X-axis so, the standard equation of the Hyperbola will be:

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

By comparing the standard parameter (transverse axis length and foci) with the given one, we get

$2a=8\Rightarrow a=4$ and $c=5$

Now, As we know the relation in a hyperbola

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

$⇒b^2=c^2-a^2$

$⇒b^2=5^2-4^2$

$⇒b^2=25-16=9$

Hence, The Equation of the hyperbola is $\frac{x^2}{16}-\frac{y^2}{9}=1$.

Question:11 Find the equations of the hyperbola satisfying the given conditions: Foci (0, ±13), the conjugate axis is of length 24.

Answer:

Given, in a hyperbola: Foci (0, ±13), the conjugate axis is of length 24.

Here, foci are on the Y-axis so, the standard equation of the Hyperbola will be:

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

By comparing the standard parameter (length of conjugate axis and foci) with the given one, we get

$2b=24\Rightarrow b=12$ and $c=13$

Now, As we know the relation in a hyperbola

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

$⇒a^2=c^2-b^2$

$⇒a^2=13^2-12^2$

$⇒a^2=169-144=25$

Hence, The Equation of the hyperbola is $\frac{y^2}{25}-\frac{x^2}{144}=1$.

Question:12 Find the equations of the hyperbola satisfying the given conditions: Foci $(\pm 3\sqrt5, 0)$, the latus rectum is of length 8.

Answer:

Given, in a hyperbola: Foci $(\pm 3\sqrt5, 0)$, the latus rectum is of length 8.

Here, foci are on the X-axis so, the standard equation of the Hyperbola will be ;

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

By comparing the standard parameter (length of latus rectum and foci) with the given one, we get

$c=3\sqrt{5}$ and
$\frac{2b^2}{a}=8$
$\Rightarrow 2b^2=8a$
$\Rightarrow b^2=4a$

Now, As we know the relation in a hyperbola

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

$⇒c^2=a^2+4a$

$⇒a^2+4a=(3\sqrt{5})^2$

$⇒a^2+4a=45$

$⇒a^2+9a-5a-45=0$

$⇒(a+9)(a-5)=0$

$⇒a=-9$ or, $a=5$

Since $a$ can never be negative,

$a=5$

$⇒a^2=25$

So, $b^2=4a=4(5)=20$

Hence, The Equation of the hyperbola is $\frac{x^2}{25}-\frac{y^2}{20}=1$.

Question:13 Find the equations of the hyperbola satisfying the given conditions: Foci (± 4, 0), the latus rectum is of length 12

Answer:

Given, in a hyperbola: Foci (± 4, 0), the latus rectum is of length 12

Here, foci are on the X-axis so, the standard equation of the Hyperbola will be:

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

By comparing the standard parameter (length of latus rectum and foci) with the given one, we get

$c=4$ and

$\frac{2b^2}{a}=12$
$\Rightarrow 2b^2=12a$
$\Rightarrow b^2=6a$

Now, As we know the relation in a hyperbola

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

$⇒c^2=a^2+6a$

$⇒a^2+6a=4^2$

$⇒a^2+6a=16$

$⇒a^2+8a-2a-16=0$

$⇒(a+8)(a-2)=0$

$⇒a=-8$ or, $a=2$

Since $a$ can never be negative,

$a=2$

$⇒a^2=4$

So, $b^2=6a=6(2)=12$

Hence, The Equation of the hyperbola is $\frac{x^2}{4}-\frac{y^2}{12}=1$.

Question:14 Find the equations of the hyperbola satisfying the given conditions: vertices (± 7,0), $e = \frac{4}{3}$

Answer:

Given, in a hyperbola: vertices (± 7,0), and $e = \frac{4}{3}$

Here, Vertices is on the X-axis so, the standard equation of the Hyperbola will be:

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

By comparing the standard parameter (Vertices and eccentricity) with the given one, we get,

$a=7$ and

$e=\frac{c}{a}$
$⇒\frac{4}{3}=\frac{c}{7}$
$⇒c=\frac{28}{3}$

Now, As we know the relation in a hyperbola

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

$⇒b^2=c^2-a^2$

$⇒b^2=\left(\frac{28}{3}\right)^2-7^2$

$⇒b^2=\left(\frac{784}{9}\right)-49$

$⇒b^2=\left(\frac{784-441}{9}\right)=\frac{343}{9}$

Hence, The Equation of the hyperbola is $\frac{x^2}{49}-\frac{9y^2}{343}=1$.

Question:15 Find the equations of the hyperbola satisfying the given conditions: Foci $(0,\pm\sqrt{10})$, passing through (2,3)

Answer:

Given, in a hyperbola: Foci $(0,\pm\sqrt{10})$, passing through (2,3)

Since the foci of the hyperbola are on the Y-axis, the equation of the hyperbola will be of the form:

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

By comparing the standard parameter (foci) with the given one, we get

$c=\sqrt{10}$

Now As we know, in a hyperbola

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

$⇒a^2+b^2=10—----(1)$

Now As the hyperbola passes through the point (2,3)

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

$⇒9b^2-4a^2=a^2b^2—------(2)$

Solving Equation (1) and (2), we get,

$9(10-a^2)-4a^2=a^2(10-a^2)$

$⇒a^4-23a^2+90=0$

$⇒(a^2)^2-18a^2-5a^2+90=0$

$⇒(a^2-18)(a^2-5)=0$

$⇒a^2=18$ or, $a=5$

Now, as we know in a hyperbola $c$ is always greater than $a$,
we choose the value $a^2=5$

So, $b^2=10-a^2=10-5=5$

Hence The Equation of the hyperbola is $\frac{y^2}{5}-\frac{x^2}{5}=1$.

Question:1 If a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus.

Answer:

Let the parabolic reflector open towards the right.

So, the equation of the parabolic reflector will be $y^2=4ax$

Now, Since this curve will pass through the point (5,10) if we assume origin at the optical centre,

So,

$10^2=4a(5)$

$⇒a=\frac{100}{20}=5$

Hence, The focus of the parabola is $(a,0)=(5,0)$.

Alternative Method:

As we know on any concave curve, focus, $f=\frac{R}{2}$

$\therefore$ Focus, $f=\frac{R}{2}=\frac{10}{2}=5$

Hence the focus is 5 cm right to the optical centre.

Question:2 An arch is in the form of a parabola with its axis vertical. The arch is 10 m high and 5 m wide at the base. How wide is it 2 m from the vertex of the parabola?

Answer:

Since the Axis of the parabola is vertical, Let the equation of the parabola be $x^2=4ay$

it can be seen that this curve will pass through the point ($\frac52$, 10) if we assume the origin at the bottom end of the parabolic arch.

So,

$\left(\frac{5}{2}\right)^2=4a(10)$

$⇒a=\frac{25}{160}=\frac{5}{32}$

Hence, the equation of the parabola is:

$x^2=4\times\frac{5}{32}\times y$

$⇒x^2=\frac{20}{32} y$

$⇒x^2=\frac{5}{8} y$

Now, when y = 2 the value of x will be:

$x=\sqrt{(\frac{5}{8}\times2)}=\sqrt{\frac{5}{4}}=\frac{\sqrt{5}}{2}$

Hence the width of the arch at this height is:

$2x=2\times\frac{\sqrt{5}}{2}=\sqrt{5}$

Question:3 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 being 6 m Find the length of a supporting wire attached to the roadway 18 m from the middle

Answer:

Given,

The width of the parabolic cable = 100m

The length of the shorter supportive wire attached = 6m

The length of the longer supportive wire attached = 30m

Since the rope opens upwards, the equation will be of the form $x^2=4ay$

Now if we consider the origin at the centre of the rope, the equation of the curve will pass through points: (50, 30-6) = (50,24)

$50^2=4a\times 24$

$⇒a=\frac{625}{24}$

Hence the equation of the parabola is

$x^2=4\times \frac{625}{24}\times y$

$⇒x^2= \frac{625}{6}\times y$

Now at a point, 18 m right from the centre of the rope, the x coordinate of that point will be 18.
So, by the equation, the y-coordinate will be:

$y=\frac{x^2}{4a}=\frac{18^2}{4\times \frac{625}{24}}\approx 3.11$ m

Hence, the length of the supporting wire attached to the roadway from the middle is
(3.11 + 6)m = 9.11 m

Question:4 An arch is in the form of a semi-ellipse. It is 8 m wide and 2 m high at the centre. Find the height of the arch at a point 1.5 m from one end.

Answer:

The equation of the semi-ellipse will be of the form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\:,y>0$

Now, According to the question,

the length of major axis = 2a = 8
$\Rightarrow a=4$

The length of the semi-major axis = 2
$\Rightarrow b=2$

Hence the equation will be,

$\frac{x^2}{4^2}+\frac{y^2}{2^2}=1\:,y>0$

$⇒\frac{x^2}{16}+\frac{y^2}{4}=1\:,y>0$

Now, at point 1.5 cm from the end, the x coordinate is $4-1.5 = 2.5$

So, the height at this point is

$\frac{(2.5)^2}{16}+\frac{y^2}{4}=1$
$\Rightarrow y=\sqrt{4(1-\frac{2.5^2}{16})}$

$⇒y\approx 1.56$ m

Hence, the height of the required point is 1.56 m.

Question:5 A rod of length 12 cm moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point P on the rod, which is 3 cm from the end in contact with the x-axis.

Answer:

Let $\theta$ be the angle that the rod makes with the ground.

Now, at a point 3 cm from the end,

$\cos\theta=\frac{x}{9}$

At the point of touching the ground

$\sin\theta=\frac{y}{3}$

Now, As we know the trigonometric identity,

$\cos^2\theta+\sin^2\theta=1$

$⇒\left (\frac{x}{9} \right )^2+\left ( \frac{y}{3} \right )^2=1$

$⇒\frac{x^2}{81}+\frac{y^2}{9}=1$

Hence, the equation is $\frac{x^2}{81}+\frac{y^2}{9}=1$

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.

Answer:

Given the parabola, $x^2=12y$

Comparing this equation with $x^2=4ay$, we get, $a=3$

Now, As we know the coordinates of the ends of the latus rectum are:

$(2a,a)$ and $(-2a,a)$

So, the coordinates of the latus rectum are,

$(2a,a)$ and $(-2a,a)=(6,3)$ and $(-6,3)$

Now the area of the triangle with coordinates (0,0),(6,3) and (-6,3)

Width of the triangle $= 2 \times 6=12$

Height of the triangle = 3

So, the area of the triangle
= $\frac{1}{2}\times$ Base $\times$ height
= $\frac{1}{2}\times12\times3$
= $18$

Hence, the required area is 18 square units.

Question:7 A man running a racecourse notes that the sum of the distances from the two flag posts from him is always 10 m and the distance between the flag posts is 8 m. Find the equation of the posts traced by the man.

Answer:

As we know if a point moves in a plane in such a way that its distance from two points remains constant then the path is an ellipse.

Now, According to the question,

the distance between the point from where the sum of the distance from a point is constant = 10
$\Rightarrow 2a=10$
$\Rightarrow a=5$

Now, the distance between the foci = 8
$\Rightarrow 2c=8$
$\Rightarrow c=4$

Now, As we know the relation,

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

$⇒b^2=a^2-c^2$

$⇒b=\sqrt{a^2-c^2}=\sqrt{5^2-4^2}=\sqrt{25-16}=\sqrt{9}=3$

Hence the equation of the ellipse is:

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

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

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

Hence the path of the man will be:

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

Question:8 An equilateral triangle is inscribed in the parabola $y^2 = 4 ax$, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.

Answer:

Given, an equilateral triangle inscribed in parabola with the equation $y^2 = 4 ax$

The one coordinate of the triangle is A(0,0).

Now, let the other two coordinates of the triangle are

$B(x,\sqrt{4ax})$ and $C(x,-\sqrt{4ax})$

Now, Since the triangle is equilateral,

$BC=AB=CA$

$⇒2\sqrt{4ax}=\sqrt{(x-0)^2+(\sqrt{4ax}-0)^2}$

$⇒x^2=12ax$

$⇒x=12a$

The coordinates of the points of the equilateral triangle are,

$(0,0),(12,\sqrt{4a\times 12a}),(12,-\sqrt{4a\times 12a})$
$=(0,0),(12,4\sqrt{3}a)$ and $(12,-4\sqrt{3}a)$

So, the side of the triangle is:

$2\sqrt{4ax}=2\times4\sqrt{3}a=8\sqrt{3}a$