Solution

Given that radius of the circle is a i.e., $(h, k)=(a, a)$

So, the equation of required circle is

$(x-a)^{2}+(y-a)^{2} =a^{2} \\ $

$\Rightarrow x^{2}-2 a x+a^{2}+y^{2}-2 a y+a^{2} =a^{2} \\ $

$\Rightarrow x^{2}+y^{2}-2 a x-2 a y+a^{2} =0 $

Solution

Given points are

$ \begin{aligned} x & =\frac{2 a t}{1+t^{2}} \text { and } y=\frac{a(1-t^{2})}{1+t^{2}} \\ x^{2}+y^{2} & =\frac{4 a^{2} t^{2}}{(1+t^{2})^{2}}+\frac{a^{2}(1-t^{2})^{2}}{(1+t^{2})^{2}} \end{aligned} $

$ \Rightarrow \quad \frac{1}{a^{2}}(x^{2}+y^{2})=\frac{4 t^{2}+1+t^{4}-2 t^{2}}{(1+t^{2})^{2}} $

$ \begin{matrix} \Rightarrow & \frac{1}{a^{2}}(x^{2}+y^{2})=\frac{t^{4}+2 t^{2}+1}{(1+t^{2})^{2}} \\ \Rightarrow & \frac{1}{a^{2}}(x^{2}+y^{2})=\frac{(1+t^{2})^{2}}{(1+t^{2})^{2}} \\ \Rightarrow & x^{2}+y^{2}=a^{2}, \text { which is a required circle. } \end{matrix} $

Solution

Let the equation of circle is

$ x^{2}+y^{2}+2 g x+2 f y=0 $

Since, this circle passes through the points $A(0,0), B(a, 0)$ and $C(0, b)$.

$ \therefore \quad a^{2}+2 a g=0 $

$ \text { and } \quad b^{2}+2 b f=0 $

From Eq. (ii), $a+2 g=0 \Rightarrow g=-a / 2$

From Eq. (iii), $b+2 f=0 \Rightarrow f=-b / 2$

Hence, the coordinates of the circle are $\frac{a}{2}, \frac{b}{2}$.

Solution

Given that, centre of the circle is $(1,2)$.

$\because \quad$ Radius $=2$

So, the equation of circle is

$\begin{aligned} & (x-1)^2+(y-2)^2 =2^2 \\ \Rightarrow & x^2-2 x+1+y^2-4 y+4 =4 \\ \Rightarrow & x^2-2 x+y^2-4 y+1 =0 \\ \Rightarrow & x^2+y^2-2 x-4 y+1 =0 \end{aligned}$

Solution

Given lines,

or

$ \begin{matrix} 3 x-4 y+4=0 \\ 6 x-8 y-7=0 \\ 3 x-4 y-7 / 2=0 \end{matrix} $

It is clear that lines (i) and (ii) parallel.

Now, distance between them i.e.,

$ d=|\frac{4+7 / 2}{\sqrt{9+16}}|=|\frac{\frac{8+7}{2}}{5}|=3 / 2 $

$\therefore$ Distance between these line $=$ Diameter of these circle

$\therefore$ Diameter of the circle $=3 / 2$

and radius of the circle $=3 / 4$

Solution

Let $a$ be the radius of the circle. Then, the coordinates of the circle are $(-a,-a)$. Now, perpendicular distance from $C$ to the line $A B=$ Radius of the circle

$ d=|\frac{-3 a+4 a+8}{\sqrt{9+16}}|=|\frac{a+8}{5}| \\ $

$\because a= \pm \frac{a+8}{5} \\ $

$\text { Taking positive sign, } a=\frac{a+8}{5} \\ $

$\Rightarrow 5 a=a+8 \\ $

$\Rightarrow 4 a=8 \Rightarrow a=2 \\ $

$\text { Taking negative sign, } a=\frac{-a-8}{5} \\ $

$\Rightarrow 5 a=-a-8 \\ $

$\Rightarrow 6 a=-8 \Rightarrow a=-4 / 3 $

So, the equation of circle is

$ \begin{aligned} & (x+2)^{2}+(y+2)^{2}=2^{2} \quad[\because a=2] \\ & \Rightarrow \quad x^{2}+4 x+4+y^{2}+4 y+4=4 \\ & \Rightarrow \quad x^{2}+y^{2}+4 x+4 y+4=0 \end{aligned} $

Solution

Given equation of the circle is

$ x^{2}+y^{2}-4 x-6 y+11=0 . $

$\therefore \quad 2 g=-4$ and $2 f=-6$

So, the centre of the circle is $(-g,-f)$ i.e., $(2,3)$.

Since, the mid-point of $A B$ is $(2,3)$.

Then,

$2=\frac{3+x_1}{2}$

$\Rightarrow$

$4=3+x_1$

$\therefore$

$x_1=1$

and

$\Rightarrow$

$3=\frac{4+y_1}{2}$

$6=4+y_1 \Rightarrow y_1=2$

So, the coordinates of other end of the diameter will be $(1,2)$.

Solution

Given that, centre of the circle is $(1,-2)$ and the circle passing through the lines

$ 3 x+y=14 $

and

$ 2 x+5 y=18 $

From Eq. (i) $y=14-3 x$ put in Eq. (ii), we get

$ \Rightarrow \quad \begin{aligned} & 2 x+70-15 x=18 \\ \Rightarrow & -13 x=-52 \Rightarrow x=4 \end{aligned} $

Now, $x=4$ put in Eq. (i), we get

$ 12+y=14 \Rightarrow y=2 $

Since, point $(4,2)$ lie on these lines also lies on the circle.

$ \begin{aligned} \therefore \quad \text { Radius of the circle } & =\sqrt{(4-1)^{2}+(2+2)^{2}} \\ & =\sqrt{9+16}=5 \end{aligned} $

Now, equation of the circle is

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

$\Rightarrow x^{2}-2 x+1+y^{2}+4 y+4 =25 \\ $

$\Rightarrow x^{2}+y^{2}-2 x+4 y-20 =0 $

Solution

Given equation of circle,

$ x^{2}+y^{2}=16 $

$\therefore$ Radius $=4$ and centre $=(0,0)$

Now, perpendicular from $(0,0)$ to line $y=\sqrt{3} x+k=$ Radius of the circle

$|\frac{0-0+k}{\sqrt{3+1}}|=4$

Since the distance from the point $(m, n)$ to the line $A x+B y+k=0$ is $d=|\frac{A m+B n+C}{A^{2}+B^{2}}|$

$ \begin{matrix} \Rightarrow & \pm \frac{k}{2} & =4 \\ \therefore & k & = \pm 8 \end{matrix} $

Solution

Given equation of the circle is

$ \begin{aligned} & x^{2}+y^{2}-6 x+12 y+15=0 \\ & \therefore \quad 2 g=-6 \Rightarrow g=-3 \\ & \text { and } \\ & 2 f=12 \Rightarrow f=6 \\ & c=15 \end{aligned} $

[since, the circles are concentric]

Let radius of the required circle $=r_1$

$\therefore \quad 2 \times$ Area of the given circle $=$ Area of the required circle

$ \begin{matrix} \Rightarrow & 2[\pi(\sqrt{30})^{2}] & =\pi_1^{2} \\ \Rightarrow & 60 & =r_1^{2} \\ \Rightarrow & r_1 & =\sqrt{60} \\ \therefore & \sqrt{g^{2}+f^{2}-c} & =\sqrt{60} \\ \Rightarrow & 9+36-c & =60 \\ \Rightarrow & c & =-15 \end{matrix} $

So, the required equation of circle is $x^{2}+y^{2}-6 x+12 y-15=0$.

Solution

Consider the equation of the ellipse is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$.

$\therefore$ Length of major axis $=2 a$

Length of minor axis $=2 b$

and length of latusrectum $=\frac{2 b^{2}}{a}$

Given that,

$ \begin{aligned} \frac{2 b^{2}}{a} & =\frac{2 b}{2} \\ a & =2 b \Rightarrow b=a / 2 \\ b^{2} & =a^{2}(1-e^{2}) \end{aligned} $

$\Rightarrow$

We know that,

$\Rightarrow$

$\frac{a}2^{2}=a^{2}(1-e^{2})$

$\Rightarrow \quad \frac{a^{2}}{4}=a^{2}(1-e^{2})$

$\Rightarrow \quad 1-e^{2}=\frac{1}{4}$

$\Rightarrow$

$e^{2}=1-\frac{1}{4}$

$e=\sqrt{\frac{3}{4}}=\sqrt{\frac{3}{2}}$

Solution

Given equation of ellipse,

$ \begin{aligned} 9 x^{2}+25 y^{2} & =225 \\ \frac{x^{2}}{25}+\frac{y^{2}}{9} & =1 \\ a=5, b & =3 \\ b^{2} & =a^{2}(1-e^{2}) \\ 9 & =25(1-e^{2}) \\ \frac{9}{25} & =1-e^{2} \\ e^{2} & =1-9 / 25 \\ =\sqrt{1-9 / 25} & =\sqrt{\frac{25-9}{25}} \\ & =\sqrt{\frac{16}{25}}=4 / 5 \end{aligned} $

$ \begin{matrix} \Rightarrow & \frac{x^{2}}{25}+\frac{y^{2}}{9} & =1 \\ \Rightarrow & a=5, b & =3 \\ & \text { We know that, } & b^{2} & =a^{2}(1-e^{2}). \\ \Rightarrow & 9 & =25(1-e^{2}). \\ \Rightarrow & \frac{9}{25} & =1-e^{2} \\ \Rightarrow & e^{2} & =1-9 / 25 \\ & \therefore & e=\sqrt{1-9 / 25} & =\sqrt{\frac{25-9}{25}} \end{matrix} $

Foci $=( \pm a e, 0)=( \pm 5 \times 4 / 5,0)=( \pm 4,0)$

Solution

Given that, eccentricity $=\frac{5}{8}$, i.e., $e=\frac{5}{8}$

Let equation of the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$,

Since the foci of this ellipse is $( \pm a e, 0)$.

$\therefore \quad$ Distance between foci $=\sqrt{(a e+a e)^{2}}$

$\Rightarrow \quad 2 \sqrt{a^2 e^2}=10 \quad[\because$ distance between its foci $=10]$

$\Rightarrow \quad \sqrt{a^{2} e^{2}}=5$

$\Rightarrow \quad a^{2} e^{2}=25$

$\Rightarrow \quad a^{2}=\frac{25 \times 64}{25}$

$\therefore \quad a=8$

We know that,

$\Rightarrow$

$b^{2}=a^{2}(1-e^{2})$

$\Rightarrow \quad b^{2}=64 \quad 1-\frac{25}{64}$

$\Rightarrow \quad b^{2}=64 \frac{64-25}{64}$

$b^{2}=39$

$\therefore \quad$ Length of latusrectum of ellipse $=\frac{2 b^{2}}{a}=2 \quad \frac{39}{8}=\frac{39}{4}$

Solution

Given that, $e=2 / 3$ and latusrectum $=5$

$ \begin{matrix} \text { i.e., } & \frac{2 b^{2}}{a}=5 \Rightarrow b^{2}=\frac{5 a}{2} \\ \text { We know that, } & \frac{b^{2}}{}=a^{2}(1-e^{2}) \\ \Rightarrow & \frac{5 a}{2}=a^{2} 1-\frac{4}{9} \\ \Rightarrow & \frac{5}{2}=\frac{5 a}{9} \Rightarrow a=9 / 2 \Rightarrow a^{2}=\frac{81}{4} \\ \Rightarrow & b^{2}=\frac{5 \times 9}{2 \times 2}=\frac{45}{4} \end{matrix} $

So, the required equation of the ellipse is $\frac{4 x^{2}}{81}+\frac{4 y^{2}}{45}=1$.

Solution

The equation of ellipse is $\frac{x^{2}}{36}+\frac{y^{2}}{20}=1$.

On comparing this equation with $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, we get

We know that,

$a=6, b=2 \sqrt{5}$

$\Rightarrow$

$b^{2}=a^{2}(1-e^{2})$

$\Rightarrow \quad \frac{20}{36}=1-e^{2}$

$20=36(1-e^{2})$

$\therefore \quad e=\sqrt{1-\frac{20}{36}}=\sqrt{\frac{16}{36}}$

$E=\frac{4}{6}=\frac{2}{3}$

Now,

$\therefore \quad \frac{a}{e}=\frac{\frac{6}{2}}{3}=\frac{6 \times 3}{2}=9$

and

$ -\frac{a}{e}=-9 $

$\therefore$ Distance between the directrices $=|9-(-9)|=18$

Solution

Given parabola is $y^{2}=8 x$

On comparing this parabola to the $y^{2}=4 a x$, we get

$ 8 x=4 a x \Rightarrow a=2 $

$\therefore$ Focal distance $=|x+a|=4$

$ \begin{matrix} \Rightarrow & |x+2| & =4 \\ \Rightarrow & x+2 & = \pm 4 \\ \Rightarrow & x & =2,-6 \\ \text { But } & x & \neq-6 \\ \text { For } x=2, & y^{2} & =8 \times 2 \\ \therefore & y^{2} & =16 \Rightarrow y= \pm 4 \end{matrix} $

So, the points are $(2,4)$ and $(2,-4)$.

Solution

Given equation of the parabola is $y^{2}=4 a x$

Let the coordinates of any point $P$ on the parabola be $(a t^{2}, 2 a t)$.

$ \begin{aligned} & \text { In } \triangle P O A, \quad \tan \theta=\frac{2 \text { at }}{a t^{2}}=\frac{2}{t} \\ & \Rightarrow \quad \tan \theta=\frac{2}{t} \Rightarrow t=2 \cot \theta \\ & \therefore \quad \quad \text { length of } O P=\sqrt{(0-a t^{2})^{2}+(0-2 a t)^{2}} \\ & =\sqrt{a^{2} t^{4}+4 a^{2} t^{2}} \\ & =a t \sqrt{t^{2}+4} \\ & =2 a \cot \theta \sqrt{4 \cot ^{2} \theta+4} \\ & =4 a \cot \theta \sqrt{1+\cot ^{2} \theta} \\ & =4 a \cot \theta \cdot cosec \theta \\ & =\frac{4 a \cos \theta}{\sin \theta} \cdot \frac{1}{\sin \theta}=\frac{4 a \cos \theta}{\sin ^{2} \theta} \end{aligned} $

Solution

Given that the coordinates, vertex of the parabola $(0,4)$ and focus of the parabola $(0,2)$.

By definition of the parabola, $\quad P B=P F$

$ \begin{matrix} \Rightarrow & & |\frac{0+y-6}{\sqrt{0+1}}| & =\sqrt{(x-0)^{2}+(y-2)^{2}} \\ \Rightarrow & & |y-6| & =\sqrt{x^{2}+y^{2}-4 y+4} \\ \Rightarrow & & x^{2}+y^{2}-4 y+4 & =y^{2}+36-12 y \\ \Rightarrow & x^{2}+8 y & =32 \end{matrix} $

Solution

Given that, line $y=m x+1$ is tangent to the parabola $y^{2}=4 x$. $\therefore$

and

$ \begin{aligned} y & =m x+1 \\ y^{2} & =4 x \end{aligned} $

From Eqs. (i) and (ii),

$ \begin{matrix} \Rightarrow & m^{2} x^{2}+2 m x+1 & =4 x \\ \Rightarrow & & m^{2} x^{2}+2 m x-4 x+1 & =0 \\ \Rightarrow & m^{2} x^{2}+x(2 m-4)+1 & =0 \\ \Rightarrow & (2 m-4)^{2}-4 m^{2} \times 1 & =0 \\ \Rightarrow & 4 m^{2}+16-16 m-4 m^{2} & =0 \\ \therefore & 16 m & =16 \\ & m & =1 \end{matrix} $

Solution

Distance between the foci i.e., and

$ \begin{aligned} 2 a e & =16 \Rightarrow a e=8 \\ e & =\sqrt{2} \\ a \sqrt{2} & =8 \\ a & =4 \sqrt{2} \\ b^{2} & =a^{2}(e^{2}-1) \\ b^{2} & =(4 \sqrt{2})^{2}[(\sqrt{2})^{2}-1] \\ & =16 \times 2(2-1) \\ & =32(2-1) \end{aligned} $

$ \text { and } $

We know that,

$\Rightarrow$

So, the equation of hyperbola is

$ \begin{aligned} & \frac{x^{2}}{32}-\frac{y^{2}}{32}=1 \\ & \Rightarrow \quad x^{2}-y^{2}=32 \end{aligned} $

Solution

Given equation of the hyperbola is

$ \begin{matrix} \Rightarrow & & 9 y^{2}-4 x^{2} & =36 \\ \Rightarrow & \frac{9 y^{2}}{36}-\frac{4 x^{2}}{36} & =\frac{36}{36} \\ \Rightarrow & \frac{y^{2}}{4}-\frac{x^{2}}{9} & =1 \\ \Rightarrow & -\frac{x^{2}}{9}+\frac{y^{2}}{4} & =1 \end{matrix} $

Since, this equation in form of $-\frac{x^{2}}{a}+\frac{y^{2}}{b^{2}}=1$, where $a=3$ and $b=2$.

$ \begin{aligned} \therefore \quad e & =\sqrt{1+\frac{a}b^{2}} \\ & =\sqrt{1+\frac{9}{4}}=\frac{\sqrt{13}}{2} \end{aligned} $

Solution

Given that eccentricity i.e., $e=3 / 2$ and $( \pm a e, 0)=( \pm 2,0)$

$ \begin{aligned} \therefore a e & =2 \\ \Rightarrow a \cdot \frac{3}{2} & =2 \Rightarrow a=4 / 3 \\ \because b^{2} & =a^{2}(e^{2}-1) \\ \Rightarrow b^{2} & =\frac{16}{9} \quad \frac{9}{4}-1 \\ \Rightarrow b^{2} & =\frac{16}{4} \quad \frac{5}{4}=+\frac{20}{9} \end{aligned} $

So, the equation of hyperbola is

$\quad \frac{x^2}{\frac{16}{9}}-\frac{y^2}{\frac{20}{9}}=1$

$\Rightarrow \frac{x^2}{4}-\frac{y^2}{5}=\frac{4}{9}$

Solution

Given lines are

and

$ \begin{aligned} 2 x-3 y-5 & =0 \\ 3 x-4 y-7 & =0 \\ \frac{x}{21-20} & =\frac{y}{-15+14}=\frac{1}{-8+9} \\ \frac{x}{1} & =\frac{y}{-1}=\frac{1}{+1} \\ x & = \pm 1, y=-1 \end{aligned} $

Since the intersection point of these lines will be coordinates of the circle i.e., coordinates of the circle as $(1,-1)$.

Let the radius of the circle is $r$.

Then

$ \pi r^{2}=154 $

$ \begin{aligned} \Rightarrow & \frac{22}{7} \times r^{2} & =154 \\ \Rightarrow & r^{2} & =\frac{154 \times 7}{22} \\ \Rightarrow & r^{2} & =\frac{14 \times 7}{2} \Rightarrow r^{2}=49 \end{aligned} $

So, the equation of circle is

$ (x-1)^{2}+(y+1)^{2}=49 $

$ \begin{matrix} \Rightarrow & x^{2}-2 x+1+y^{2}+2 y+1=49 \\ \Rightarrow & x^{2}+y^{2}-2 x+2 y=47 \end{matrix} $

Solution

Let the general equation of the circle is

$ x^{2}+y^{2}+2 g x+2 f y+c=0 $

Since, this circle passes through the points $(2,3)$ and $(4,5)$.

$ \begin{matrix} \therefore & 4+9+4 g+6 f+c & =0 \\ \Rightarrow & 4 g+6 f+c & =-13 \\ \text { and } & 16+25+8 g+10 f+c & =0 \\ \Rightarrow & 8 g+10 f+c & =-41 \end{matrix} $

Since, the centre of the circle $(-g,-f)$ lies on the straight line $y-4 x+3=0$

i.e., $\quad+4 g-f+3=0$

From Eq. (iv), $4 g=f-3$

On putting $4 g=f-3$ in Eq. (ii), we get

$ f-3+6 f+c=-13 $

$ \Rightarrow \quad 7 f+c=10 $

From Eqs. (ii) and (iii),

$ \begin{gathered} 8 g+12 f+2 c=-26 \\ 8 g+10 f+c=-41 \\ -\quad-\quad+ \\ \hline 2 f+c=15 \end{gathered} $

From Eqs. (ii) and (vi),

$ \begin{aligned} & 7 f+c=-10 \\ & 2 f+c=15 \\ & \frac{–}{5 f=-25} \\ & \therefore \quad f=-5 \\ & \text { Now, } \quad c=10+15=25 \\ & \text { From Eq. (iv), } \quad 4 g+5+3=0 \\ & \Rightarrow \quad g=-2 \\ & \text { From Eq. (i), equation of the circle is } x^{2}+y^{2}-4 x-10 y+25=0 \text {. } \end{aligned} $

Solution

Given centre of the circle is $(3,-1)$.

Now,

$ O P=|\frac{6+5+18}{\sqrt{4+25}}|=\frac{29}{\sqrt{29}}=\sqrt{29} $

$\ln \triangle O P B$

$ \begin{matrix} O B^{2}=O P^{2}+P B^{2} & {[\because A B=6 \Rightarrow P B=3]} \\ O B^{2}=29+9 \Rightarrow O B^{2}=38 \end{matrix} $

$\Rightarrow$

So, the radius of circle is $\sqrt{38}$,

$\therefore$ Equation of the circle with radius $r=\sqrt{38}$ and centre $(3,-1)$ is

$\Rightarrow \quad(x-3)^{2}+(y+1)^{2}=38$

$\Rightarrow \quad x^{2}-6 x+9+y^{2}+2 y+1=38$

$x^{2}+y^{2}-6 x+2 y=28$

Solution

Let the coordinates of centre of the required circle are $(h, k)$, then the centre of another circle is $(1,2)$

Conic Sections

So, it is clear that $P$ is the mid-point of $C_1 C_2$.

$\begin{matrix} \therefore & 5=\frac{1+h}{2} \Rightarrow h=9 \\ \text { and } & 5=\frac{2+k}{2} \Rightarrow k=8\end{matrix} $

So, the equation of and required circle is

$(x-9)^{2}+(y-8)^{2} =25 \\ $

$\Rightarrow x^{2}-18 x+81+y^{2}-16 y+64 =25 \\ $

$\Rightarrow x^{2}+y^{2}-18 x-16 y+120 =0 $

Solution

Let equation of circle be

$(x-h)^{2}+(y-k)^{2}=r^{2} \\ $

$\Rightarrow (x-h)^{2}+(y-k)^{2}=9 $

$ y=x-1 i . e ., k=h-1 $

Now, the circle passes through the point $(7,3)$.

$ \begin{matrix} \therefore & (7-h)^{2}+(3-k)^{3}=9 \\ \Rightarrow & 49-14 h+h^{2}+9-6 k+k^{2}=9 \\ \Rightarrow & h^{2}+k^{2}-14 h-6 k+49=0 \end{matrix} $

On putting $k=h-1$ in Eq. (iii), we get

$ h^{2}+(h-1)^{2}-14 h-6(h-1)+49=0 $

$ \begin{matrix} \Rightarrow & h^{2}+h^{2}-2 h+1-14 h-6 h+6+49=0 \\ \Rightarrow & 2 h^{2}-22 h+56=0 \\ \Rightarrow & h^{2}-11 h+28=0 \\ \Rightarrow & h^{2}-7 h-4 h+28=0 \\ \Rightarrow & h(h-7)-4(h-7)=0 \end{matrix} $

$\Rightarrow$

$ (h-7)(h-4)=0 $

$\therefore$

When

$\therefore$ Centre $(7,6)$

When $\quad h=4$, then $k=3$

$\therefore$ Centre $-(4,3)$

So, the equation of circle when centre $(7,6)$, is

$ \begin{matrix} \Rightarrow & x^{2}-14 x+49+y^{2}-12 y+36=9 \\ \Rightarrow & x^{2}+y^{2}-14 x-12 y+76=0 \end{matrix} $

$ (x-7)^{2}+(y-6)^{2}=9 $

When centre $(4,3)$, then the equation of the circle is

$ (x-4)^{2}+(y-3)^{2}=9 $

$\Rightarrow$

$ x^{2}-8 x+16+y^{2}-6 y+9=9 $

$\Rightarrow$

$ x^{2}+y^{2}-8 x-6 y+16=0 $

Solution

(i) Given that, directrix $=0$ and focus $=(6,0)$

So, the equation of the parabola

$ (x-6)^{2}+y^{2}=x^{2} $

$ \begin{matrix} \Rightarrow & x^{2}+36-12 x+y^{2}=x^{2} \\ \Rightarrow & y^{2}-12 x+36=0 \end{matrix} $

(ii) Given that, vertex $=(0,4)$ and focus $=(0,2)$

So, the equation of parabola is

$\sqrt{(x-0)^{2}+(y-2)^{2}} =|y-6| \\ $

$\Rightarrow x^{2}+y^{2}-4 y+4 =y^{2}-12 y+36 \\ $

$\Rightarrow x^{2}-4 y+12 y-32 =0 \\ $

$\Rightarrow x^{2}+8 y-32 =0 \\ $

$\Rightarrow x^{2} =32-8 y $

(iii) Given that, focus at $(-1,-2)$ and directrix $x-2 y+3=0$

So, the equation of parabola is $\sqrt{(x+1)^{2}+(y+2)^{2}}=|\frac{x-2 y+3}{\sqrt{1+4}}|$

$ \begin{matrix} \Rightarrow & x^{2}+2 x+1+y^{2}+4 y+4=\frac{1}{5}[x^{2}+4 y^{2}+9-4 x y-12 y+6 x] \\ \Rightarrow & 4 x^{2}+4 x y+y^{2}+4 x+32 y+16=0 \end{matrix} $

Solution

Let the coordinates of the point be $(x, y)$, then according to the question,

$\sqrt{(x-3)^{2}+y^{2}}+\sqrt{(x-9)^{2}+y^{2}}=12 \\ $

$\Rightarrow \sqrt{(x-3)^{2}+y^{2}}=12-\sqrt{(x-9)^{2}+y^{2}} $

Conic Sections

On squaring both sides, we get

$ x^{2}-6 x+9+y^{2} =144+(x^{2}-18 x+81+y^{2})-24 \sqrt{(x-9)^{2}+y^{2}} \\ $

$\Rightarrow 12 x-216 =-24 \sqrt{(x-9)^{2}+y^{2}} \\ $

$\Rightarrow x-18 =-2 \sqrt{(x-9)^{2}+y^{2}} \\ $

$\Rightarrow x^{2}-36 x+324 =4(x^{2}-18 x+81+y^{2}) \\ $

$3 x^{2}+4 y^{2}-36 x =0 $

Solution

Let the point be $P(x, y)$.

$\therefore$ Distance from $(0,4)=\sqrt{x^{2}+(y-4)^{2}}$

So, the distance from the line $y=9$ is $|\frac{y-9}{\sqrt{1}}|$

$ \begin{matrix} \therefore & \sqrt{x^{2}+(y-4)^{2}}=\frac{2}{3}|\frac{y-9}{1}| \\ \Rightarrow & x^{2}+y^{2}-8 y+10=\frac{4}{9}(y^{2}-18 y+81) \\ \Rightarrow & 9 x^{2}+9 y^{2}-72 y+144=4 y^{2}-72 y+324 \\ \Rightarrow & 9 x^{2}+5 y^{2}=180 \end{matrix} $

Solution

Let the points be $P(x, y)$

$\therefore$ Distance of $P$ from $(4,0) \sqrt{(x-4)^{2}+y^{2}}$

and the distance of $P$ from $(-4,0) \sqrt{(x+4)^{2}+y^{2}}$

Now,

$ \begin{aligned} & \sqrt{(x+4)^{2}+y^{2}}-\sqrt{(x-4)^{2}+y^{2}}=2 \\ & \sqrt{(x+4)^{2}+y^{2}}=2+\sqrt{(x-4)^{2}+y^{2}} \end{aligned} $

$\Rightarrow$

On squaring both sides, we get

$x^{2}+8 x+16+y^{2} =4+x^{2}-8 x+16+y^{2}+4 \sqrt{(x-4)^{2}+y^{2}} \\ $

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

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

$\Rightarrow 16 x^{2}-8 x+1 =x^{2}+16-8 x+y^{2} \\ $

$15x^{2}-y^{2} =15 \text { which is a parabola. } $

Solution

(i) Given that, vertices $=( \pm 5,0)$, foci $=( \pm 7,0)$ and $a= \pm 5$

$ \begin{matrix} \therefore & ( \pm a e, 0)=( \pm 7,0) \\ \text { Now } & a e=7 \Rightarrow 5 e=7 \\ \Rightarrow & e=7 / 5 \\ \because & b^{2}=a^{2}(e^{2}-1) \\ \Rightarrow & b^{2}=25 \frac{49}{25}-1 \\ \Rightarrow & b^{2}=25 \frac{49-25}{25} \\ \Rightarrow & b^{2}=24 \end{matrix} $

So,the equation of parabola is

$ \frac{x^{2}}{25}-\frac{y^{2}}{24}=1 \quad[\because a^{2}=25 \text { and } b^{2}=24] $

(ii) Vertices $=(0, \pm 7), e=4 / 3$

$ \begin{matrix} \therefore & b & =7, e=4 / 3 \\ & \ddots & e^{2} & =1+\frac{a^{2}}{b^{2}} \\ \Rightarrow & & \frac{16}{9}-1 & =\frac{a^{2}}{49} \\ \Rightarrow & & \frac{7}{9} & =\frac{a^{2}}{49} \Rightarrow a^{2}=\frac{343}{9} \end{matrix} $

So, the equation of hyperbola is

$-\frac{x^{2} \times 9}{343}+\frac{y^{2}}{49} =1 \\ $

$\Rightarrow -\frac{9 x^{2}}{7}+y^{2} =49 \\ $

$\Rightarrow 9 x^{2}-7 y^{2}+343 =0 $

(iii) Given that, foci $=(0, \pm \sqrt{10})$

$ \begin{aligned} \because & b e & =\sqrt{10} \\ \Rightarrow & a^{2}+b^{2} & =10 \\ \Rightarrow & a^{2} & =10-b^{2} \end{aligned} $

$\therefore$ Equation of the hyperbola be

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

Since, this hyperbola passes through the point $(2,3)$.

$ \begin{matrix} \therefore & -\frac{4}{a^{2}}+\frac{9}{b^{2}}=1 \\ \Rightarrow & \frac{-4}{10-b^{2}}+\frac{9}{b^{2}}=1 \end{matrix} $

$ \begin{matrix} \Rightarrow & \frac{-4 b^{2}+90-9 b^{2}}{b^{2}(10-b^{2})}=1 \\ \Rightarrow & -13 b^{2}+90=10 b^{2}+b^{4} \\ \Rightarrow & b^{4}-23 b^{2}+90=0 \\ \Rightarrow & b^{4}-18 b^{2}-5 b^{2}+90=0 \\ \Rightarrow & b^{2}(b^{2}-18)-5(b^{2}-18)=0 \\ \Rightarrow & (b^{2}-18)(b^{2}-5)=0 \\ \Rightarrow & b^{2}=18 \Rightarrow b= \pm 3 \sqrt{2} \\ \text { or } & b^{2}=5 \Rightarrow b=\sqrt{5} \\ \therefore & b^{2}=18 \text { then } a^{2}=-8 \\ \text { When } & a^{2}=5 \text {, then } b^{2}=5 \end{matrix} $

$ \therefore \quad b^{2}=18 \text { then } a^{2}=-8 \quad \text { [not possible] } $

So, the equation of hyperbola is

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

$\Rightarrow y^{2}-x^{2} =5 $

Solution

False

Given equation of the circle is

$ \begin{matrix} \therefore & x^{2}+y^{2}+6 x+2 y & =0 \\ \text { Since given line is } x+3 y=0 . & & \text { Centre }=(-3,-1) \\ \Rightarrow \quad & -3-3 & \neq 0 \end{matrix} $

So, this line is not diameter of the circle.

Solution

False

Given circle is $x^{2}+y^{2}-14 x-10 y-151=0$.

$\therefore \quad$ Centre $=(7,5)$

and $\quad$ Radius $=\sqrt{49+25+151}=\sqrt{225}=15$

So, the distance between the point $(2,-7)$ and centre of the circle is given by

$\therefore$ Shortest distance, $d=|13-15|=2$

$ \begin{aligned} d_1 & =\sqrt{(2-7)^{2}+(-7-5)^{2}} \\ & =\sqrt{25+144}=\sqrt{169}=13 \end{aligned} $

Solution

True

Given circle is $x^{2}+y^{2}=a^{2}$

$\therefore$ Radius of circle $=a$ and centre $=(0,0)$

$\therefore$ Distance from point $(l, m)$ and centre is $\sqrt{(0-e)^{2}+(0-m)^{2}}=a$

$\Rightarrow \quad l^{2}+m^{2}=a^{2}$

So, $l, m$ lie on the circle.

Solution

False

Given circle is $S \equiv x^{2}+y^{2}-2 x+6 y+1=0$.

Since, the point is $(1,2)$.

Now,

$ \begin{gathered} S_1 \equiv 1+4-2+12+1 \\ S_1>0 \end{gathered} $

$\Rightarrow$

So, the $(1,2)$ lies outside the circle.

Solution

True

Given equation of a line is

and

$l x+m y+n=0 \\ $

$\text { parabola } y^{2}=4 a x $

From Eq. (i), $x=-\frac{m y+n}{l}$ put in Eq. (ii), we get

$y^{2} =-\frac{4 a(m y+n)}{l} \\ $

$\Rightarrow l y^{2} =-4 a m y-4 a x \\ $

$\Rightarrow l y^{2}+4 a m y+4 a n =0 \\ $

$\text { For tangent, } D =0 \\ $

$\Rightarrow 16 a^{2} m^{2} =4 l \times 4 a n \\ $

$\Rightarrow 16 a^{2} m^{2} =16 \text { anl } \\ $

$\Rightarrow a m^{2} =n l $

Solution

False

Given equation of the ellipse is $\frac{x}{16}+\frac{y^{2}}{25}=1$.

which is in form of $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, where $b>a$

$ \begin{aligned} & \therefore \quad \text { Foci, } S=(0, b e), S^{\prime}(0,-b e) \\ & \therefore \quad e=\sqrt{1-\frac{a^{2}}{b^{2}}} \\ & =\sqrt{\frac{25-16}{25}}=3 / 5 \end{aligned} $

$ \text { Foci, } S=0, \frac{3 \times 5}{5}, S^{\prime}=0,-\frac{3 \times 5}{5} \text { i.e., } S=(0,3), S^{\prime}=(0,-3) $

Let the coordinate of point $P$ be $(x, y)$ then $P S+P S^{\prime}=2 b=2 \times 5=10$

Solution

True

Given equation of line is

and

$ \begin{aligned} 2 x+3 y & =12 \\ \text { ellipse } \frac{x^{2}}{9}+\frac{y^{2}}{4} & =2 \end{aligned} $

Since, the equation of tangent at $(x_1, y_1)$ is $\frac{x x_1}{9}+\frac{y y_1}{4}=2$.

$\therefore$ Tangent at $(3,2)$,

$ \begin{aligned} & & \frac{3 x}{9}+\frac{2 y}{4}=2 \\ \Rightarrow & \frac{x}{3}+\frac{y}{2} & =2 \\ \Rightarrow & 2 x+3 y & =12 \text {, which is a given line. } \end{aligned} $

Hence, the statement is true.

Solution

True

Given equations of line are

and

From Eq. (i),

$ \begin{aligned} \sqrt{3} x-y-4 \sqrt{3} k & =0 \\ \sqrt{3} k x+k y-4 \sqrt{3} & =0 \\ 4 \sqrt{3} k & =\sqrt{3} x-y \end{aligned} $

$ \begin{matrix} \Rightarrow & k=\frac{\sqrt{3} x-y}{4 \sqrt{3}} \text { put in Eq. (ii), we get } \\ \Rightarrow & \sqrt{3} x \frac{\sqrt{3} x-y}{4 \sqrt{3}}+\frac{\sqrt{3} x-y}{4 \sqrt{3}} y-4 \sqrt{3}=0 \\ \Rightarrow & \frac{1}{4}(\sqrt{3} x^{2}-x y)+\frac{1}{4} x y-\frac{y^{2}}{\sqrt{3}}-4 \sqrt{3}=0 \\ \Rightarrow & \frac{\sqrt{3}}{4} x^{2}-\frac{y^{2}}{4 \sqrt{3}}-4 \sqrt{3}=0 \\ \Rightarrow & 3 x^{2}-y^{2}-48=0 \\ \Rightarrow & 3 x^{2}=48, \text { which is a hyperbola. } \end{matrix} $

Solution

The perpendicular distance from centre $(3,-4)$ to the line is, $d=|\frac{\mid 5-48-12}{\sqrt{25+144}}|=\frac{45}{13}$

So, the required equations of the circle is $(x-3)^{2}+(y+4)^{2}=\frac{45}13^{2}$.

Solution

Given equations of line are

From Eqs. (i) and (ii),

$ \begin{aligned} y & =x+2 \\ 3 y & =4 x \\ 2 y & =3 x \end{aligned} $

$\Rightarrow$

$ \frac{4 x}{3}=x+2 $

$\Rightarrow x=6$

On putting $x=6$ in Eq. (i), we get

$ \begin{aligned} & y & =8 \\ \therefore & \text { Point, } A & =(6,8) \end{aligned} $

From Eqs. (i) and (iii),

$ \begin{aligned} & \frac{3 x}{2}=x+2 \\ & 3 x=2 x+4 \Rightarrow x=4 \end{aligned} $

When

$\therefore$

From Eqs. (ii) and (iii)

Now,

Let the equation of circle is

$ \begin{aligned} x & =4, \text { then } y=6 \\ \text { Point, } B & =(4,6) \\ x_1 & =0_1, y=0 \\ C & =(0,0) \end{aligned} $

$ x^{2}+y^{2}+2 g x n+2 f y+c=0 $

Since, the points $A(6,8), B(4,6)$ and $C(0,0)$ lie on this circle.

$ 36+64+12 g+16 f+c=0 $

$ \begin{aligned} \Rightarrow & 12 g+16 f+c & =-100 \\ \text { and } & 16+36+8 g+12 f+c & =0 \\ \Rightarrow & 8 g+12 f+c & =-52 \\ \Rightarrow & c & =0 \end{aligned} $

From Eqs. (iv), (v) and (vi),

$ \begin{aligned} \Rightarrow & 12 g+16 f & =-100 \\ \Rightarrow & 3 g+4 f+25 & =0 \\ \Rightarrow & \frac{g}{+52-75} & =\frac{f}{50-39}=\frac{1}{9-8} \\ \Rightarrow & \frac{g}{-23} & =\frac{f}{11}=\frac{1}{1} \\ \Rightarrow & g & =-23, f=11 \end{aligned} $

So, the equation of circle is

$ \Rightarrow \quad \begin{aligned} x^{2}+y^{2}-46 x+22 y+0 & =0 \\ x^{2}+y^{2}-46 x+22 y & =0 \end{aligned} $

Solution

Let equation of the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$.

$ \begin{aligned} & \therefore \quad 29=6 \text { and } 2 b=4 \\ & \Rightarrow \\ & a=3 \text { and } b=2 \end{aligned} $

We know that,

$ \begin{aligned} c^{2} & =a^{2}-b^{2}=(3)^{2}-(2)^{2} \\ & =9-4=5 \Rightarrow c=\sqrt{5} \end{aligned} $

$ \begin{aligned} \therefore \quad \text { Length of string } & =A C^{\prime}+C^{\prime} C+A C \\ & =a+c+2 c+a c \\ & =2 a+2 c=6+2 \sqrt{5} \end{aligned} $

$\therefore \quad$ Distances between the pins $=2 \sqrt{5}=C c^{\prime}$

Solution

Given that, foci of the ellipse are $(0, \pm b e)$.

$\because \quad$ be $\equiv 1$

$\therefore \quad$ Length of minor axis, $2 a=1 \Rightarrow a=1 / 2$

$ \begin{matrix} e^{2} & =1-\frac{a^{2}}{b^{2}} \\ \Rightarrow & (b e)^{2} & =b^{2}-a^{2} \Rightarrow 1=b^{2}-\frac{1}{4} \\ \Rightarrow & 1+\frac{1}{4} & =b^{2} \Rightarrow \frac{5}{4}=b^{2} \end{matrix} $

So, the equation of ellipse is

$ \frac{x^{2}}{\frac{1}{4}}+\frac{y^{2}}{5 / 4}=1 \Rightarrow \frac{4 x^{2}}{1}+\frac{4 y^{2}}{5}=1 $

Solution

Given that, focus at $F(-1,-2)$ and directrix is $x-2 y+3=0$

Let any point on the parabola be $(x, y)$.

$ \begin{aligned} & \therefore \quad P F=|\frac{x-2 y+3}{\sqrt{1+4}}| \\ & \Rightarrow \quad(x+1)^{2}+(y+2)^{2}=\frac{(x-2 y+3)^{2}}{5} \\ & \Rightarrow \quad 5[x^{2}+2 x+1+y^{2}+4 y+4]=x^{2}+4 y^{2}+9-4 x y-12 y+6 x \\ & \Rightarrow \quad 4 x^{2}+y^{2}+4 x+32 y+16=0 \end{aligned} $

Solution

Let the equation of the hyperbola be $-\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$.

Then vertices $=(0, \pm b)=(0, \pm 6)$

$\therefore b =6 \text { and } e=5 / 3 \\ $

$\because e =\sqrt{1+\frac{a^{2}}{b^{2}}} \Rightarrow \frac{25}{9}=1+\frac{a^{2}}{36} \\ $

$\Rightarrow \frac{25-9}{9} =\frac{a^{2}}{36} \Rightarrow 16=\frac{a^{2}}{4} \Rightarrow a^{2}=48 $

So, the equation of hyperbola is,

$ \begin{aligned} \frac{-x^{2}}{48}+\frac{y^{2}}{36} & =1 \Rightarrow \frac{y^{2}}{36}-\frac{x^{2}}{48}=1 \\ \because \quad \text { Foci } & =(0, \pm b e)=0, \pm \frac{5}{3} \times 6=(0, \pm 10) \end{aligned} $

Solution

(c) Given that, centre of the circle is $(1,2)$.

$\because \quad C P=\sqrt{9+16}=5=$ Radius of the circle

$\therefore \quad$ Required area $=\pi r^{2}=25 \pi$

Solution

(c) Let centre of the circle be $(a, a)$, then equation of the circle is $(x-a)^{2}+(y-a)^{2}=a^{2}$.

Since, the point $(3,6)$ lies on this circle, then

$(3-a)^{2}+(6-a)^{2} =a^{2} \\ $

$ \Rightarrow a^{2}+9-6 a+36-12 a+a^{2} =a^{2} \\ $

$ \Rightarrow a^{2}-18 a+45 =0 \\ $

$ \Rightarrow a^{2}-15 a-3 a+45 =0 \\ $

$ \Rightarrow a(a-15)-3(a-15) =0 \\ $

$ \Rightarrow (a-3)(a-15) =0 \\ $

$ \Rightarrow a =3, a=15 $

So, the equation of circle is

$(x-3)^{2}+(y-3)^{2} =9 \\ $

$\Rightarrow x^{2}-6 x+9+y^{2}-6 y+9 =9 \\ $

$\Rightarrow x^{2}+y^{2}-6 x-6 y+9 =0 $

Solution

(c) Let general equation of the circle is $x^{2}+y^{2}+2 g h+2 f y+c=0$.

Since the point $(0,0)$ and $(2,3)$ lie on it $c=0$.

$ \begin{matrix} \therefore & 4+9+4 g & +6 f=0 \\ \Rightarrow & 2 g & +3 f=-13 / 2 \\ \text { Since the centre lie on } Y \text {-axis, then } g & =0 . \\ \therefore & 3 f & =-13 / 2 \\ \Rightarrow & f & =-13 / 6 \end{matrix} $

So, the equation of circle is

$ \Rightarrow \quad 6 x^{2}+6 y^{2}-13 y=0 $

Solution

(c) Given that, length of the median $A D=3 a$

$\because \quad$ Radius of the circle $=\frac{3}{2} \times$ Length of median

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

So, the equation of the circle is $x^{2}+y^{2}=4 a^{2}$.

Solution

(a) Given that, focus of parabola at $F(0,-3)$ and equation of directrix is $y=3$.

Let any point on the parabola is $P(x, y)$.

Then,

$ P F=|y-3| $

$ \begin{matrix} \Rightarrow & & \sqrt{(x-0)^{2}+(y+3)^{2}} & =|y-3| \\ \Rightarrow & x^{2}+y^{2}+6 y+9 & =y^{2}-6 y+9 \\ & \Rightarrow & x^{2}+12 y & =0 \\ \Rightarrow & x^{2} & =-12 y \end{matrix} $

Solution

(b) Given that, parabola is

$ y^{2}=4 a x $

$\therefore$ Length of latusrectum $=4 a$

Since, the parabola passes through the point $(3,2)$.

Then,

$ \begin{gathered} 4=4 a(3) \\ a=1 / 3 \\ 4 a=4 / 3 \end{gathered} $

$ \begin{aligned} & \Rightarrow \\ & \therefore \end{aligned} $

Solution

(a) Here, vertex $=(-3,0)$

$\therefore a=-3$ and directrix, $x+5=0$

Since, axis of the parabola is a line perpendicular to directrix and $A$ is the mid-point of AS.

$ \begin{matrix} \text { Then, } & -3 & =\frac{x_1-5}{2} \\ \Rightarrow & -6 & =x_1-5 \Rightarrow x_1=-1, \\ & \therefore & 0 & =\frac{0+y_1}{2} \Rightarrow y_1=0 \\ & \therefore & S & =(-1,0) \\ \Rightarrow & P M & =P S \\ \Rightarrow & & x+5 & =\sqrt{(x+1)^{2}+y^{2}} \\ \Rightarrow & x^{2}+2 x+1+y^{2} & =x^{2}+10 x+25 \\ \Rightarrow & y^{2} & =+8 x+24 \\ & & y^{2} & =+8(x+3) \end{matrix} $

Solution

(a) Given that, focus of the ellipse is $(1,-1)$ and the equation of directrix is $x-y-3=0$ and $e=\frac{1}{2}$

Let $P(x, y)$ and $F(1,-1)$.

$\therefore \quad \frac{P F}{\text { Distance of } P \text { from }(x-y-3=0)}=\frac{1}{2}$

$ \begin{matrix} \Rightarrow & \frac{\sqrt{(x-1)^{2}+(y+1)^{2}}}{\frac{|x-y-3|}{\sqrt{2}}}=\frac{1}{2} \\ \Rightarrow & \frac{2[x^{2}-2 x+1+y^{2}+2 y+1]}{(x-y-3)^{2}}=\frac{1}{4} \\ \Rightarrow & 8 x^{2}-16 x+16+8 y^{2}+16 y=x^{2}+y^{2}+9-2 x y+6 y-6 x \\ \Rightarrow & 7 x^{2}+7 y^{2}+2 x y-10 x+10 y+7=0 \end{matrix} $

Solution

(d) Given equation of ellipse is

$ 3 x^{2}+y^{2}=12 $

$ \begin{aligned} & \Rightarrow \quad \frac{x^{2}}{4}+\frac{y^{2}}{12}=1 \\ & \therefore \quad a^{2}=4 \Rightarrow a=2 \\ & \text { and } \quad b^{2}=12 \Rightarrow b=2 \sqrt{3} \\ & \because \quad b>a \\ & \therefore \quad \text { Length of latusrectum }=\frac{2 \times 4}{2 \sqrt{3}}=\frac{4}{\sqrt{3}}=\frac{2 a^{2}}{b} \end{aligned} $

Solution

(b) Given that,

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

We know that,

$ e=\sqrt{1-\frac{a^{2}}{b^{2}}} \Rightarrow e^{2}=\frac{(b^{2}-a^{2})}{b^{2}} $

$ \begin{matrix} \Rightarrow & b^{2} e^{2}=b^{2}-a^{2} \\ \Rightarrow & a^{2}=b^{2}(1-e^{2}) \end{matrix} $

Solution

(c) Length of latusrectum of the hyperbola i.e.,

$ 8=\frac{2 b^{2}}{a} \Rightarrow b^{2}=4 a $

$\therefore$ Distance between the foci $=2 a e$

Since, transverse axis be $a$ and conjugate axis be $b$.

$ \begin{matrix} \therefore & \frac{1}{2}(2 a e)=2 b \\ \Rightarrow & a e=2 b \\ \Rightarrow & b^{2}=a^{2}(e^{2}-1) \end{matrix} $

From Eqs. (i) and (ii),

$4 a =\frac{a^{2} e^{2}}{4} \\ $

$ \Rightarrow 16 a =a^{2} e^{2} \\ $

$\because 16 =a e^{2} \Rightarrow a=\frac{16}{e^{2}} \\ $

$\Rightarrow 4 a =a^{2}(e^{2}-1) \\ $

$\Rightarrow \frac{4}{a} =e^{2}-1 \\ $

$\Rightarrow \frac{4 e^{2}}{16} =e^{2}-1 $

$ \begin{aligned} \Rightarrow & e^{2} 1-\frac{4}{16} & =1 \\ \Rightarrow & e^{2} \frac{12}{16} & =1 \Rightarrow e^{2}=\frac{16}{12} \\ \Rightarrow & e^{2} & =\frac{4}{3} \Rightarrow e=\frac{2}{\sqrt{3}} \end{aligned} $

Solution

(a) Given that, distance between the foci of hyperbola

$ \begin{matrix} \text { i.e., } & 2 a e & =16 \Rightarrow a e=8 \\ \text { and } & e & =\sqrt{2} \\ \text { Now, } & \sqrt{2} a & =8 \\ \Rightarrow & a & =4 \sqrt{2} \\ \because & b^{2} & =a^{2}(e^{2}-1) \\ \Rightarrow & b^{2} & =32(2-1) \\ \Rightarrow & b^{2} & =32 \\ & \therefore & \frac{x^{2}}{32}-\frac{y^{2}}{32} & =1 \\ \Rightarrow & & x^{2}-y^{2} & =32 \end{matrix} $

Solution

(a) Given that, eccentricity of the hyperbola, $e=3 / 2$

and

$\because$

$\Rightarrow$

$\because$

$\Rightarrow$

$\Rightarrow \quad b^{2}=\frac{20}{9}$

So, the equation of the hyperbola is

$ \frac{x^{2}}{\frac{16}{9}}-\frac{y^{2}}{20 / 9}=1 \Rightarrow \frac{x^{2}}{4}-\frac{y^{2}}{5}=\frac{4}{9} $

$ \begin{aligned} \text { foci } & =( \pm 2,0),( \pm a e, 0) \\ a e & =2 \\ a \times 3 / 2 & =2 \Rightarrow a=4 / 3 \\ b^{2} & =a^{2}(e^{2}-1) \\ b^{2} & =\frac{16}{9} \frac{9}{4}-1 \Rightarrow b^{2}=\frac{16}{9} \frac{5}{4} \\ b^{2} & =\frac{20}{9} \end{aligned} $