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} $

Thinking Process

General equation of the circle passing through the origin is $x^{2}+y^{2}+2 y x+2 f y=0$.

Now, satisfied the given points to get the values of $g$ and $f$. The centre of the circle is $(-g,-f)$.

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

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

Thinking Process

The distance between two parallel lines $a x+b y+c_1=0$ and $a x+b y+c_2=0$ is given by,

i.e., $d=|\frac{c_1-c_2}{\sqrt{a^{2}+b^{2}}}|$. Use this formula to solve the above problem.

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} $

Thinking Process

First of all get the centre of the circle from the given equation, then find the mid-point of the diameter of the circle.

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,

$\Rightarrow$

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

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

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

$\Rightarrow$

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

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

Thinking Process

Find the values of $a$ and $b$ by the given equation of ellipse, then use the formula $b^{2}=a^{2}(1-e^{2})$ to get the value of $e$.

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 $=\frac{2 b^{2}}{a}=2 \quad \frac{39}{8}=\frac{39}{4}$

Thinking Process

First of all find the values of $a$ and $b$ using the formula $b^{2}=a^{2}(1-e^{2})$, then get the equation of the ellipse.

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} $

Thinking Process

First of all find the value of $a$ and $b$ using the given condition, then put them in $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ to get the required equation of the hyperbola.

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

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

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

Thinking Process

First of all find the intersection point of the given lines, then get radius of circle from given area. Now, use formula equation of circle with centre $(h, k)$ and radius a is $(x-h)^{2}+$ $(y-k)^{2}=a^{2}$.

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 $

Thinking Process

First of all let the equation of a circle with centre $(h, k)$ and radius $r$ is

$(x-h)^{2}+(y-h)^{2}=r^{2}$, then we get the value of $(h, k)$ using given condition.

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 $

Thinking Process

Consider the points $(x, y)$, and apply the condition given in the problem, then get the set of all points.

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 $

Thinking Process

If a line is the diameter of circle, then the centre of the circle should lies on line. Use this property to solve the given problem.

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.

Thinking Process

If the $x_1, y_1$ lies inside the circle $S \equiv x^{2}+y^{2}+2 g x+2 f y+c=0$, then $x_1^{2}+y_1^{2}+2 g x_1+2 f y_1+c<0$ and it $S>0$, then the point lies outside 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.

Thinking Process

First of all eliminate $k$ from the given equations of line, then get the equation of hyperbola.

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 $=2 \sqrt{5}=C c^{\prime}$

Thinking Process

First of all get the value of $a$ and $b$ with the help of given condition in the problem, then we get the required equation of the ellipse.

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} $

Thinking Process

First of all find the value of $a$ and $b$ from the given equation, after that get length of latusrectum by using formula $\frac{2 a^{2}}{b}$.

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} $

Thinking Process

The distance between the foci of hyperbola is $2 a$ e and $b^{2}=a^{2}(e^{2}-1)$. Use this relation to set the value of $a$ and $b$.

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} $