Conic Section

Let point $P(-2,y)$ is point on parabola from def. of parabola

$PS=PM$

$\Rightarrow {(y-1)}^2=\frac{{(-4+y+2)}^2}{5}$

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

$\Rightarrow 4y^2-6y+1=0$

Sum of roots $=\frac{6}{4}=\frac{3}{2}$

\(\begin{aligned}
& x-y+2=0 \\
& x+y+2=0
\end{aligned}\)

\( P Q=8 \)
\(\Rightarrow(a-b)^2+(a+2+b+2)^2=64 \)
\((a-b)^2+(a+b+4)=64\)
\(2 a^2+2 b^2-2 a b+16+2(a b+4 b+4 a)=64 \)
\(\Rightarrow 2 a^2+2 b^2+8 a+8 b=48 \)
\(\Rightarrow a^2+b^2+4 a+4 b-24=0\)
\(\Rightarrow (a+2)^2+(b+2)^2=32\)
\begin{aligned}
& h=\frac{2 b+1(a)}{3}, k=\frac{2(-2-b)+1(a+2)}{3} \\
& 3 h=2 b+a, 3 k=a-2-2 b \rightarrow \text { Solve for } a \text { and } b . \\
& a=\frac{3 h+3 k+2}{2}, b=\frac{3 h-3 k-2}{4} \\
& (a+2)=\left(\frac{3 h+3 k+6}{2}\right), b+2=\left(\frac{3 h-3 k+6}{2}\right) \\
& \Rightarrow\left(\frac{3 h+3 k+6}{2}\right)^2+\left(\frac{3 h-3 k+6}{2}\right)^2=32 \\
& 9(x+y+2)^2+9(x-y+2)^2=128 \\
& 18\left[x^2+y^2+4 x+4\right]=128 \Rightarrow x^2+y^2+4 x-\frac{28}{9}=0
\end{aligned}\)

$H:\frac{y^2}{b^2}-\frac{x^2}{a^2}=1,e=\sqrt{3}$

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

$a^2=2b^2$

length of L.R. \(=\frac{2 a^2}{b}=4 \sqrt{3}\)

\(P(\alpha, 6)\) lie on \(\frac{y^2}{3}-\frac{x^2}{6}=1\)

$12-\frac{{\alpha }^2}{6}=1\Rightarrow {\alpha }^2=66$

Foci \(=(0, \pm b e)=(0,3), (0,-3)\)

Let $d_1, d_2$ be focal distances of $P(\alpha ,6)$

$d_1=\sqrt{{\alpha }^2+(6+be{)}^2},d_2=\sqrt{{\alpha }^2+(6-be{)}^2}$

$d_1=\sqrt{66+81},d_2=\sqrt{66+9}$

$\beta =d_1{d}_2=\sqrt{147\times 75}=105$

${\alpha }^2+\beta =66+105=171$

$H:\frac{y^2}{b^2}-\frac{x^2}{a^2}=1,e=\sqrt{3}$

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

$a^2=2b^2$

length of L.R. \(=\frac{2 a^2}{b}=4 \sqrt{3}\)

\(P(\alpha, 6)\) lie on \(\frac{y^2}{3}-\frac{x^2}{6}=1\)

$12-\frac{{\alpha }^2}{6}=1\Rightarrow {\alpha }^2=66$

Foci \(=(0, \pm b e)=(0,3), (0,-3)\)

Let $d_1, d_2$ be focal distances of $P(\alpha ,6)$

$d_1=\sqrt{{\alpha }^2+(6+be{)}^2},d_2=\sqrt{{\alpha }^2+(6-be{)}^2}$

$d_1=\sqrt{66+81},d_2=\sqrt{66+9}$

$\beta =d_1{d}_2=\sqrt{147\times 75}=105$

${\alpha }^2+\beta =66+105=171$

Point of contact are $\left(\frac{\mp a^{2}m}{\sqrt{a^2{m}^2+b^2}},\frac{\pm b^2}{\sqrt{a^2{m}^2+b^2}}\right)$

$\mathrm{A}\left(\frac{-16}{5},\frac{9}{5}\right) \mathrm{and} \mathrm{B}\left(\frac{16}{5},\frac{-9}{5}\right)$
Point \(D\) is $\left(\frac{12}{5},\frac{12}{5}\right)$
Area of $ABD=\frac{1}{2}\left|\begin{array}{ccc}-\frac{16}{5}& \frac{9}{5}& 1\\ \frac{16}{5}& \frac{-9}{5}& 1\\ \frac{12}{5}& \frac{12}{5}& 1\end{array}\right|=12$

Area of $ABCD$ is $=24$

Length of minor axis \(=2 b=\frac{1}{4}(2 a e)\)

$\Rightarrow 4b=ae\Rightarrow 16b^2=a^2{e}^2$

$\Rightarrow 16\left(a^2-a^2{e}^2\right)=a^2{e}^2\Rightarrow 16a^2=17a^2{e}^2$

$\Rightarrow e^2=\frac{16}{17}\Rightarrow e=\frac{4}{\sqrt{17}}$

$f\left(x\right)=x^2+9, g\left(x\right)=\frac{x}{x-9}$

$a=f\left(g\left(10\right)\right)=f\left(\frac{10}{10-9}\right)$

$=f\left(10\right)=109$

$b=g\left(f\left(3\right)\right)=g\left(9+9\right)$

$=g\left(18\right)=\frac{18}{9}=2$

$E:\frac{x^2}{109}+\frac{y^2}{2}=1$

${\mathrm{e}}^2=1-\frac{2}{109}=\frac{107}{109}$

$\ell =\frac{2\left(2\right)}{\sqrt{109}}=\frac{4}{\sqrt{109}}$

$8{\mathrm{e}}^2+{\ell }^2=\frac{8\left(107\right)}{109}+\frac{16}{109}$

$=8$

$f\left(x\right)=x^2+9, g\left(x\right)=\frac{x}{x-9}$

$a=f\left(g\left(10\right)\right)=f\left(\frac{10}{10-9}\right)$

$=f\left(10\right)=109$

$b=g\left(f\left(3\right)\right)=g\left(9+9\right)$

$=g\left(18\right)=\frac{18}{9}=2$

$E:\frac{x^2}{109}+\frac{y^2}{2}=1$

${\mathrm{e}}^2=1-\frac{2}{109}=\frac{107}{109}$

$\ell =\frac{2\left(2\right)}{\sqrt{109}}=\frac{4}{\sqrt{109}}$

$8{\mathrm{e}}^2+{\ell }^2=\frac{8\left(107\right)}{109}+\frac{16}{109}$

$=8$

$P(a{t}^2,2at)$

$\Rightarrow P(4t^2,8t)$

$=(1,-4)$

$\Rightarrow t=\frac{-1}{2}$;$Q\left(\frac{a}{t^2},\frac{-2a}{t}\right)$

$S(4,0)$ is the focus & $PS=a+a^2$

$QS=a+\frac{a}{t^2}$

$\frac{PS}{QS}=t^2=\frac{4}{1}=\frac{m^{}}{n}$

$m^2+n^2=17$

For \(y=x^2\), slope \(=2x\).

\(2x=4\Rightarrow x=2\), so \(P=(2,4)\).

For \(x^2+y^2=2\), slope \(=-\dfrac{x}{y}\).

\(-\dfrac{x}{y}=-1\Rightarrow x=y\).

Since \(x^2+y^2=2\), we get \(2x^2=2\Rightarrow x=1\), so \(Q=(1,1)\).

For \(x^2+4y^2=8\), slope \(=-\dfrac{x}{4y}\).

\(-\dfrac{x}{4y}=-\dfrac12\Rightarrow x=2y\).

\((2y)^2+4y^2=8\Rightarrow 8y^2=8\Rightarrow y=1\), so \(R=(2,1)\).

Let centre of circle be \(C(h,k)\).

Since \(P=(2,4)\) and \(R=(2,1)\), perpendicular bisector of \(PR\) is \(y=\dfrac52\).

So centre is \(C(h,\dfrac52)\).

Using \(CQ=CR\):

\((h-1)^2+\left(\dfrac52-1\right)^2=(h-2)^2+\left(\dfrac52-1\right)^2\)

\((h-1)^2=(h-2)^2\Rightarrow h=\dfrac32\)

Radius \(=CR=\sqrt{\left(\dfrac32-2\right)^2+\left(\dfrac52-1\right)^2}\)

\(r=\sqrt{\dfrac14+\dfrac94}=\sqrt{\dfrac{10}{4}}=\sqrt{\dfrac52}\)

$\left(\frac{\mp a^{2}m}{\sqrt{a^2{m}^2+b^2}},\frac{\pm b^2}{\sqrt{a^2{m}^2+b^2}}\right)$

$A\left(\frac{-16}{5},\frac{9}{5}\right)B\left(\frac{16}{5},\frac{-9}{5}\right)$

Point D is $\left(\frac{12}{5},\frac{12}{5}\right)$

Area of $ABD=\frac{1}{2}\left|\begin{array}{ccc}-\frac{16}{5}& \frac{9}{5}& 1\\ \frac{16}{5}& \frac{-9}{5}& 1\\ \frac{12}{5}& \frac{12}{5}& 1\end{array}\right|$

= 12

Area of ABCD is 24

$$\begin{gathered} \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\text{ foci are }(\pm a{e}_1,0) \\ \frac{x^2}{A^2}-\frac{y^2}{B^2}=1\text{ foci are }\left(\pm A{e}_2,0\right) \\ \Rightarrow 2{\mathrm{ae}}_1=2\sqrt{3}\Rightarrow {\mathrm{ae}}_1=\sqrt{3} \\ \text{ and }2A{e}_2=2\sqrt{3}\Rightarrow A{e}_2=\sqrt{3} \\ \text{and} 2{\mathrm{ae}}_1=2A{e}_2\Rightarrow \frac{e_1}{e_2}=\frac{A}{a}=\frac{1}{3}\left(\text{Given}\right) \\ \Rightarrow a=3A ..\text{.}\left(1\right) \\ \text{ Now }a-A=2\Rightarrow a-\frac{a}{3}=2 \\ \Rightarrow a=3\text{ and }A=1 \\ a{e}_1=\sqrt{3}\Rightarrow {\mathrm{e}}_1=\frac{1}{\sqrt{3}}\text{ and }{\mathrm{e}}_2=\sqrt{3} \\ {\mathrm{b}}^2={\mathrm{a}}^2\left(1-{{\mathrm{e}}_1}^2\right) \\ {\mathrm{b}}^2=6 \\ \text{ and }{\mathrm{B}}^2={\mathrm{A}}^2\left({\left({\mathrm{e}}_2\right)}^2-1\right)=\left(2\right)\Rightarrow {\mathrm{B}}^2=2 \\ \text{sum of }\mathrm{LR}=\frac{2{\mathrm{b}}^2}{\mathrm{a}}+\frac{2{\mathrm{B}}^2}{\mathrm{A}}=8 \end{gathered}$$

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

It passes through \((6, 4\sqrt{3})\).

\(\Rightarrow \frac{36}{a^2}-\frac{48}{b^2}=1\)\(\ldots(1)\)

Also, \(15 \mathrm{e}^2-34 \mathrm{e}+15=0\)

\(\Rightarrow 15 \mathrm{e}^2-25 \mathrm{e}-9 \mathrm{e}+15=0\)

\(\mathrm{e}=\frac{5}{3}\) or \(\frac{3}{5} \Rightarrow \mathrm{e}=\frac{5}{3}\)

\(\Rightarrow \mathrm{e}^{2}=\frac{25}{9}\)

\(1+\frac{b^2}{a^2}=\frac{25}{9} \Rightarrow \frac{b^2}{a^2}=\frac{16}{9}\) \(\ldots(2)\)

Using (1) and (2):

\(\Rightarrow \frac{36}{a^2}-\frac{48}{16 a^2} \times 9=1 \Rightarrow a=3, b=4\)

Now, the length of the latus rectum of the hyperbola \(\frac{x^2}{b^2}-\frac{y^2}{2\left(a^2+1\right)}=1\) is:

\(=\frac{4\left(a^2+1\right)}{b}=\frac{4(10)}{4}=10\)

Parametric point \(P\) on \(x^2=4 y\) is \(P\left(2 t, t^2\right)\)

∴ mirror image of \(P\) in \(x-y=1\) is

$Q\equiv \left(2t-\frac{2\cdot 1\cdot \left(2t-t^2-1\right)}{2},t^2+\frac{2\cdot 1\cdot \left(2t-t^2-1\right)}{2}\right)$

$Q\equiv \left(t^2+1,2t-1\right)\equiv (h,k)$

So, \(h=t^2+1, k=2t-1\)

\(\Rightarrow t=\frac{k+1}{2}\)

\(\Rightarrow h= \frac{(k+1)^2}{4} +1\)

∴ locus of \(Q\) is $x=\frac{{(y+1)}^2}{4}+1$ which is the required parabola.

$\therefore {(y+1)}^2=4\left(x-1\right)$

$\therefore a=1,b=4,c=1$

$\therefore a+b+c=6$

Given \( \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \)

passes \((4,3) \)
\( \frac{16}{a^2}+\frac{9}{b^2}=1 \ldots \)(1)
\(e=\frac{\sqrt{5}}{3} \)
\( e^2=\frac{5}{9}\)
\( 1-\frac{a^2}{b^2}=\frac{5}{9} \)
\( \frac{a^2}{b^2}=\frac{4}{9} \ldots\)(2)

From (1) and (2)

\( \frac{16}{a^2}+\frac{4}{a^2}=1\)
\( \frac{20}{a^2}=1 \Rightarrow a^2=20 \ \& \ b^2=45\)
\(\ell(L R)=\frac{2 a^2}{b}=\frac{2 \times 20}{\sqrt{45}}=\frac{40}{\sqrt{45}}\)
\( =\frac{40 \sqrt{5}}{3 \sqrt{5} \times \sqrt{5}}=\frac{8 \sqrt{5}}{3}\)

parabola passes through \((1,-1)\) so

\( -1=1+p+q \)
\( p+q=-2 \)
\( q=-2-p \ldots\)(1)

Distance from \(x\)-axis

$\left|\frac{-D}{4a}\right|=\left|\frac{-\left(p^2-4q\right)}{4\left(1\right)}\right|=\left|\frac{4q-p^2}{4}\right|$
$=\left|\frac{4(-2-p)-p^2}{4}\right|$ from (1)
$=\left|\frac{-8-4p-p^2}{4}\right|=\left|\frac{-\left(p^2+4p+8\right)}{4}\right|=\left|\frac{-\left((p+2{)}^2+4\right)}{4}\right|$

Minimum at \(p=-2 \Rightarrow q=0\)

\(p^2+q^2=4+0=4\)

Given: $\frac{x^2}{25}+\frac{y^2}{16}=1$

Equation of chord with given middle point

$$\begin{gathered} \mathrm{T}={\mathrm{S}}_1 \\ \Rightarrow \frac{3\mathrm{x}}{25}+\frac{\mathrm{y}}{16}-1=\frac{9}{25}+\frac{1}{16}-1 \end{gathered}$$

On simplifying, we get

$$\begin{gathered} 48\mathrm{x} + 25\mathrm{y} = 144 + 25 \\ 48\mathrm{x} + 25\mathrm{y} = 169 \end{gathered}$$

Let the roots be \(\alpha\) and \(2 \alpha\)

Sum of roots: \(3 \alpha=\frac{-9(k-1)}{k^2-15 k+27}\)

\(\Rightarrow \alpha=\frac{-3(k-1)}{k^2-15 k+27}\)

Product of roots: \(2 \alpha^2=\frac{18}{k^2-15 k+27}\)

\(\Rightarrow \alpha^2=\frac{9}{k^2-15 k+27}\)

Now, substituting the expression for \(\alpha\) into the equation for \(\alpha^2\):

\(\left(\frac{-3(k-1)}{k^2-15 k+27}\right)^2=\frac{9}{k^2-15 k+27}\)

\((k-1)^2=k^2-15 k+27\)

\(k^2-2 k+1=k^2-15 k+27\)

\(\Rightarrow k=2\)

Now, length of Latus Rectum \(=6 k=12\)

$$\begin{gathered} \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\text{ foci are }(\pm a{e}_1,0) \\ \frac{x^2}{A^2}-\frac{y^2}{B^2}=1\text{ foci are }\left(\pm A{e}_2,0\right) \\ \Rightarrow 2{\mathrm{ae}}_1=2\sqrt{3}\Rightarrow {\mathrm{ae}}_1=\sqrt{3} \\ \text{ and }2A{e}_2=2\sqrt{3}\Rightarrow A{e}_2=\sqrt{3} \\ \text{and} 2{\mathrm{ae}}_1=2A{e}_2\Rightarrow \frac{e_1}{e_2}=\frac{A}{a}=\frac{1}{3}\left(\text{Given}\right) \\ \Rightarrow a=3A ..\text{.}\left(1\right) \\ \text{ Now }a-A=2\Rightarrow a-\frac{a}{3}=2 \\ \Rightarrow a=3\text{ and }A=1 \\ a{e}_1=\sqrt{3}\Rightarrow {\mathrm{e}}_1=\frac{1}{\sqrt{3}}\text{ and }{\mathrm{e}}_2=\sqrt{3} \\ {\mathrm{b}}^2={\mathrm{a}}^2\left(1-{{\mathrm{e}}_1}^2\right) \\ {\mathrm{b}}^2=6 \\ \text{ and }{\mathrm{B}}^2={\mathrm{A}}^2\left({\left({\mathrm{e}}_2\right)}^2-1\right)=\left(2\right)\Rightarrow {\mathrm{B}}^2=2 \\ \text{sum of }\mathrm{LR}=\frac{2{\mathrm{b}}^2}{\mathrm{a}}+\frac{2{\mathrm{B}}^2}{\mathrm{A}}=8 \end{gathered}$$

\(a e=\sqrt{10}\) and \(\frac{a}{e}=\frac{9}{\sqrt{10}}\)

\(\Rightarrow a^2=9\) and \(e=\frac{\sqrt{10}}{3}\)

Now, ${\left(ae\right)}^2=a^2+b^2$

$10=9+b^2\Rightarrow b^2=1$

$l=\frac{2b^2}{a}=\frac{2\left(1\right)}{3}$

$\Rightarrow 9\left({\text{e}}^2+l\right)$

$=9\left(\frac{10}{9}+\frac{2}{3}\right)$

\(= 10 + 6 = 16\)

\( R(0,-3) \quad \mathrm{P}(\alpha, 0) \quad Q(\beta, 0) \)

\( r=\sqrt{5} \)
\( (x+1)^2+(y+1)^2=5 \)
\( (\alpha+1)^2+1=5 \)
\( (\alpha+1)^2=4 \)
\( \alpha=1,-3 \quad \text {Similarly for} \beta \)
\( P(1,0), Q(-3,0) \)
\( \text { Area }=\frac{1}{2} \times 4 \times 3=6\)

$$\begin{gathered} f\left(t\right)=\frac{-3}{4}+2t-t^2 \\ =\frac{-3}{4}-\left(t^2-2t+1-1\right) \\ =\frac{-3}{4}-{\left(t-1\right)}^2+1 \\ =\frac{1}{4}-{\left(t-1\right)}^2 \end{gathered}$$

maximum value of

$f\left(t\right)=\frac{1}{4}=e$

$\Rightarrow e^2=\frac{1}{16}\Rightarrow \frac{a^2-b^2}{a^2}=\frac{1}{16}$ .... \((i)\)

$\because \frac{2b^2}{a}=30\Rightarrow b^2=15a$ .... \((ii)\)

from \((i)\) & \((ii)\)

$16\left(a^2-15a\right)=a^2$

$\Rightarrow 15a^2-16\times 15a=0$

$a=16$, $b^2=240$

Hence, $a^2+b^2=256+240=496$

$$\begin{gathered} \text{ The given equation of hyperbola can be re-written as, } \\ \frac{x^2}{{\left(\frac{1}{2}\right)}^2}-\frac{y^2}{{\left(\frac{1}{3}\right)}^2}=1 \\ \text{So, }a=\frac{1}{2}, b=\frac{1}{3} \\ \text{then, }e=\sqrt{1+\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}} \\ e=\sqrt{1+\frac{4}{9}} \\ e=\frac{\sqrt{13}}{3} \\ \therefore \text{ Focii }=(\pm \mathrm{ae},0)=\left(\pm \frac{\sqrt{13}}{6},0\right) \end{gathered}$$

$$\begin{gathered} \text{Given,} \\ \left(\sqrt{3},\frac{1}{2}\right)\text{ lies on }\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1, \mathrm{then} \\ \frac{3}{{\mathrm{a}}^2}+\frac{1}{4{\mathrm{b}}^2}=1 \\ \frac{3}{a^2}+\frac{1}{4\left(a^2\right)\left(1-e^2\right)}=1 \\ 12\left(1-{\mathrm{e}}^2\right)+1=4{\mathrm{a}}^2\left(1-{\mathrm{e}}^2\right) ........\left(i\right) \\ \text{We know that,} \\ \text{focal distance }=(a\pm e{x}_1) \\ \text{Now, Product of focal distances } \\ =\left(a+e{x}_1\right)\left(a-e{x}_1\right) \\ \frac{7}{4}=a^2-e^2{x}_1^2 \\ Put x_1= \sqrt{3} \\ \frac{7}{4}=a^2-e^2\left(3\right) \\ \Rightarrow a^2=\frac{7}{4}+3e^2 \\ \Rightarrow 4{\mathrm{a}}^2=7+12{\mathrm{e}}^2 ..........\left(\mathrm{ii}\right) \\ \mathrm{Now}, \\ \mathrm{On} \mathrm{solving} \mathrm{eqn}. \left(\mathrm{i}\right) \mathrm{and} \left(\mathrm{ii}\right), \mathrm{we} \mathrm{get} \\ \Rightarrow 12{\mathrm{e}}^4-17{\mathrm{e}}^2+6=0 \\ \therefore {\mathrm{e}}^2=\frac{17\pm \sqrt{289-288}}{24}=\frac{3}{4}, \frac{2}{3} \\ \therefore \mathrm{e}=\frac{\sqrt{3}}{2}, \sqrt{\frac{2}{3}} \\ \therefore \text{ Difference of eccentricity }=\frac{\sqrt{3}}{2}-\sqrt{\frac{2}{3}} \\ =\frac{3-2\sqrt{2}}{2\sqrt{3}} \end{gathered}$$

$$\begin{gathered} \text{ We know that the focus of the} \\ \text{parabola is }(h+1,k) \\ \text{ It is given that focus lies between } \\ x+y=1\text{ and }x+y=3 \\ \text{i.e.} \\ 1<x+y<3 \\ 1<\mathrm{h}+1+\mathrm{k}<3 \\ 0<\mathrm{h}+\mathrm{k}<2 \end{gathered}$$

Given that distance between the foci of $E_1$ is 4

$$\begin{gathered} \Rightarrow 2\mathrm{ae}=4 \\ \Rightarrow 2\mathrm{a}\left(\frac{1}{\sqrt{3}}\right)=4 \\ \Rightarrow a=2\sqrt{3} \end{gathered}$$

$$\begin{gathered} \text{Also,} e^2=1-{\left(\frac{b}{a}\right)}^2 \\ \Rightarrow \frac{1}{3}=1-\frac{b^2}{12} \\ \Rightarrow b^2=8 \end{gathered}$$

Next product of their lengths of latus rectums is $\frac{32}{\sqrt{3}}$

$$\begin{gathered} \frac{2{ \mathrm{b}}^2}{\mathrm{a}}·\frac{2{ \mathrm{A}}^2}{ \mathrm{B}}=\frac{32}{\sqrt{3}} \\ \Rightarrow 2\left(\frac{8}{2\sqrt{3}}\right)\frac{2{ \mathrm{A}}^2}{ \mathrm{B}}=\frac{32}{\sqrt{3}} \\ \Rightarrow {\mathrm{A}}^2=2 \mathrm{B} \\ 1-\frac{{\mathrm{A}}^2}{{ \mathrm{B}}^2}=\frac{1}{3}\Rightarrow 1-\frac{2 \mathrm{B}}{{ \mathrm{B}}^2}=\frac{1}{3}\Rightarrow \mathrm{B}=3 \\ \mathrm{we} \mathrm{get} {\mathrm{A}}^2=6 \end{gathered}$$

Thus we get equation of ellipses are

$$\begin{gathered} \frac{x^2}{12}+\frac{y^2}{8}=1\dots .\left(i\right) \\ \text{and} \frac{x^2}{6}+\frac{y^2}{9}=1\dots ..\left(ii\right) \\ \text{On solving we get points of intersection as} \\ (x,y)\equiv \left(\frac{\sqrt{6}}{\sqrt{5}},\frac{6}{\sqrt{5}}\right),\left(\frac{-\sqrt{6}}{\sqrt{5}},\frac{6}{\sqrt{5}}\right), \\ \left(\frac{\sqrt{6}}{\sqrt{5}},\frac{-6}{\sqrt{5}}\right),\left(\frac{-\sqrt{6}}{\sqrt{5}},\frac{-6}{\sqrt{5}}\right) \\ \text{Area of rectangle=}\frac{24\sqrt{6}}{5} \end{gathered}$$

The distance between the foci is

$2c=14-(-12)=26 \Rightarrow c=13.$

The center of the hyperbola is the midpoint of the foci,

which is $\left(\frac{1+1}{2},\frac{14+(-12)}{2}\right)=(1,1)$

Since the foci have the same x-coordinate,

the hyperbola has a vertical transverse axis.

The equation is of the form $\frac{(y-1{)}^2}{a^2}-\frac{(x-1{)}^2}{b^2}=1$

The distance from the center (1, 1) to the point (1, 6)

on the hyperbola is $a=6-1=5.$

We know that $c^2=a^2+b^2$

$$\begin{gathered} {13}^2=5^2+b^2 \\ \Rightarrow b = 12 \end{gathered}$$

The length of the latus rectum is given by

$\frac{2b^2}{a}=\frac{2\left(144\right)}{5}=\frac{288}{5}.$

For domain of \(f(x)\)

$$\begin{gathered} {\log }_5\left({\log }_7\left(9x-x^2-13\right)\right)>0 \\ \Rightarrow 9x-x^2-13>7 \\ \Rightarrow x^2-9x+20<0 \\ \Rightarrow \left(x-5\right)\left(x-4\right)<0 \\ \Rightarrow 4<x<5 \\ m=4, n=5 \end{gathered}$$

eccenticity of hyperbola

$$\begin{gathered} e=\sqrt{1+\frac{b^2}{a^2}}=\frac{5}{3} \\ \Rightarrow \frac{b^2}{a^2}=\frac{16}{9} \\ \Rightarrow \frac{b}{a}=\frac{4}{3} \\ \mathrm{And} \frac{2b^2}{a}=\frac{8m}{3} \\ \Rightarrow \frac{2b^2}{a}=\frac{32}{3} \\ \Rightarrow 2b^2=\frac{32}{3}\times \frac{3b}{4} \\ \Rightarrow b=4,a=3 \\ {b}^2-a^2=16-9=7 \end{gathered}$$

$$\begin{gathered} \text{Given,} \\ \text{Vertex of parabola} \left(\frac{3}{2},3\right) \\ \text{Equation of directrix: }x+2y=0 \\ \text{Now, } \\ \mathrm{Equation} \mathrm{of} \mathrm{axis} \\ \mathrm{y}-3=2\left(\mathrm{x}-\frac{3}{2}\right) \\ y-2 x=0 \\ \text{foot of directrix} \text{is } (0, 0) \\ \text{Focus} = (3, 6) \\ \text{Now the required equation of parabola} \\ (x-3{)}^2+(y-6{)}^2={\left(\frac{x+2y}{\sqrt{5}}\right)}^2 \\ 4x^2+y^2-4xy-30x-60y+225=0 \\ \text{Compare with given equation:} \\ \alpha x^2+\beta y^2-\gamma xy-30x-60y+225=0 \\ \Rightarrow \alpha =4,\beta =1,\gamma =4 \\ \Rightarrow \alpha +\beta +\gamma =9 \end{gathered}$$

$$\begin{gathered} \frac{x^2}{9}-\frac{8x-x^2}{4}=1 \\ 4x^2-72x+9x^2=36 \\ 13x^2-72x-36=0 \\ x = 6 or x=\frac{-6}{13}\left(\mathrm{rejected}\right) \\ \mathrm{now}, \frac{y^2}{4}=3 \Rightarrow y=\pm \sqrt{12} \\ \mathrm{A}(6,\sqrt{12}),\mathrm{B}(6,-\sqrt{12}),\mathrm{C}\left(\mathrm{t},\frac{2\mathrm{t}+4}{3}\right) \\ 3 \mathrm{h}=12+\mathrm{t} \Rightarrow \mathrm{t}=3\mathrm{h}-12 \\ \mathrm{and} 3\mathrm{k}=\frac{2\mathrm{t}+4}{3} \Rightarrow 2\mathrm{t}+4=9\mathrm{k} \\ \Rightarrow 2\left(3\mathrm{h}-12\right)+4=9\mathrm{k} \\ \Rightarrow 6h-24+4=9\mathrm{k} \\ \Rightarrow 6\mathrm{h}-9\mathrm{k}-20=0 \\ \Rightarrow 6\mathrm{x}-9\mathrm{y}-20=0 \end{gathered}$$

Equation of chord whose mid point ( 3, 1 )

using $T=S_1$ :
\(\begin{aligned}
& \frac{3 x}{25}+\frac{y}{16}=\frac{9}{25}+\frac{1}{16} \\
& 48 x+25 y=169 \mathrm{}
\end{aligned}\)

For domain of \(f(x)\)

$$\begin{gathered} {\log }_5\left({\log }_7\left(9x-x^2-13\right)\right)>0 \\ \Rightarrow 9x-x^2-13>7 \\ \Rightarrow x^2-9x+20<0 \\ \Rightarrow \left(x-5\right)\left(x-4\right)<0 \\ \Rightarrow 4<x<5 \\ m=4, n=5 \end{gathered}$$

eccenticity of hyperbola

$$\begin{gathered} e=\sqrt{1+\frac{b^2}{a^2}}=\frac{5}{3} \\ \Rightarrow \frac{b^2}{a^2}=\frac{16}{9} \\ \Rightarrow \frac{b}{a}=\frac{4}{3} \\ \mathrm{And} \frac{2b^2}{a}=\frac{8m}{3} \\ \Rightarrow \frac{2b^2}{a}=\frac{32}{3} \\ \Rightarrow 2b^2=\frac{32}{3}\times \frac{3b}{4} \\ \Rightarrow b=4,a=3 \\ {b}^2-a^2=16-9=7 \end{gathered}$$

For \(y^2=8 x\), we have \(4 a=8\), so \(a=2\)

Directrix is \(x=-2\), so \(A=(-2,0)\)

Let \(B=\left(2 t^2, 4 t\right)\)

Slope of \(A B=\frac{4 t-0}{2 t^2-(-2)}=\frac{4 t}{2 t^2+2}\)

\(\frac{4 t}{2 t^2+2}=\frac{3}{5} \)

\(t=3, \frac{1}{3}\)

Since \(\alpha=2 t^2>1, t=3\)

So \(B=(18,12)\)

Since \(B C\) is a focal chord, parameters of \(B\) and \(C\) satisfy \(t_1 t_2=-1\),

so parameter of \(C=-\frac{1}{3}\)

\(C=\left(2\left(-\frac{1}{3}\right)^2, 4\left(-\frac{1}{3}\right)\right)=\left(\frac{2}{9},-\frac{4}{3}\right)\)

Area of \(\triangle A B C=\left|\frac{1}{2}\left|\begin{array}{ccc}-2 & 0 & 1 \\ 18 & 12 & 1 \\ \frac{2}{9} & -\frac{4}{3} & 1\end{array}\right|\right|\)

Area \(=\frac{1}{2}\left|-\frac{160}{3}\right|=\frac{80}{3}\)

Six times area \(=6 \times \frac{80}{3}=160\)

Given line: $y=\frac{1-\alpha x}{2}$

Put this in equation of hyperbola

$\therefore x^2-9{\left(\frac{1-\alpha x}{2}\right)}^2=9$

$\Rightarrow \left(4-9{\alpha }^2\right)x^2+18\alpha x-45=0$

$\because$ line does not intersect hyperbola

$\therefore D<0$

$$\begin{gathered} \Rightarrow 324{\alpha }^2+180\left(4-9{\alpha }^2\right)<0 \\ \Rightarrow 720-1296{\alpha }^2<0 \end{gathered}$$

$\Rightarrow {\alpha }^2-\frac{5}{9}>0$

$\Rightarrow \alpha \in \left(-\infty ,-\frac{\sqrt{5}}{3}\right)\cup \left(\frac{\sqrt{5}}{3},\infty \right)$

Here $\frac{\sqrt{5}}{3}\simeq 0.74$

Hence $\alpha =0.8$

$$\begin{gathered} \frac{x^2}{9}-\frac{8x-x^2}{4}=1 \\ 4x^2-72x+9x^2=36 \\ 13x^2-72x-36=0 \\ x = 6 or x=\frac{-6}{13}\left(\mathrm{rejected}\right) \\ \mathrm{now}, \frac{y^2}{4}=3 \Rightarrow y=\pm \sqrt{12} \\ \mathrm{A}(6,\sqrt{12}),\mathrm{B}(6,-\sqrt{12}),\mathrm{C}\left(\mathrm{t},\frac{2\mathrm{t}+4}{3}\right) \\ 3 \mathrm{h}=12+\mathrm{t} \Rightarrow \mathrm{t}=3\mathrm{h}-12 \\ \mathrm{and} 3\mathrm{k}=\frac{2\mathrm{t}+4}{3} \Rightarrow 2\mathrm{t}+4=9\mathrm{k} \\ \Rightarrow 2\left(3\mathrm{h}-12\right)+4=9\mathrm{k} \\ \Rightarrow 6h-24+4=9\mathrm{k} \\ \Rightarrow 6\mathrm{h}-9\mathrm{k}-20=0 \\ \Rightarrow 6\mathrm{x}-9\mathrm{y}-20=0 \end{gathered}$$

$LR=\frac{2b^2}{a}=10$

$\Rightarrow 5a=b^2$ ...(1)

$f\left(t\right)=t^2+t+\frac{11}{12}$

$\frac{df\left(t\right)}{dt}=2t+1=0$

$\Rightarrow t=\frac{-1}{2}$

Min. value of $f\left(t\right)={\left(\frac{-1}{2}\right)}^2+\left(\frac{-1}{2}\right)+\frac{11}{12}$

$=\frac{1}{4}\frac{-1}{2}+\frac{11}{12}$

$=\frac{3-6+11}{12}=\frac{8}{12}$

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

$e^2=\frac{1-b^2}{a^2}\Rightarrow \frac{4}{9}=\frac{1-b^2}{a^2}$

$\Rightarrow \frac{b^2}{a^2}=\frac{1-4}{a}=\frac{5}{a}$

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

From (1) & (2)

$5a=\frac{5a^2}{a}\Rightarrow a=9,$

$b=\sqrt{45}=3\sqrt{5}$

$\therefore a^2+b^2=81+45=126$

For ellipse $\frac{x^2}{25}+\frac{y^2}{b^2}=1$ (b < 5)

Let $e_1$ is eccentricity of ellipse

$\therefore b^2=25(1-e_1^2)\mathrm{.....}\left(1\right)$

Again for hyperbola

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

Let $e_2$ is eccentricity of hyperbola.

$\therefore b^2=16(e_2^2-1)$ .....(2)

By (1) & (2)

$25(1-e_1^2)=16(e_2^2-1)$

Now $e_1\cdot e_2=1$ (given)

$\therefore 25(1-e_1^2)=16\left(\frac{1-e_1^2}{e_1^2}\right)$

or $e_1=\frac{4}{5}\therefore e_2=\frac{5}{4}$

Now distance between foci is 2ae

$\therefore$ Distance for ellipse $2\times 5\times \frac{4}{5}=8=\alpha$

Distance for hyperbola $=2\times 4\times \frac{5}{4}=10=\beta$

$\therefore (\alpha ,\beta )\equiv (8,10)$

Given Distance between Focl of ellipse \(=2\sqrt {3}\)

\(
\begin{array}{r}
\Rightarrow \quad 2 a_1 e_1=2 \sqrt{3} \\
\Rightarrow a_1 e_1=\sqrt{3}
\end{array}
\)

and for hyperbola \(\Rightarrow \quad 2 a_2 e_2=2 \sqrt{3}\)

\(
\begin{aligned}
& a_2 e_2=\sqrt{3} \\
& \text { and } \frac{e_1}{e_2}=\frac{1}{3} \text { and } a_1-a_2=2 \\
& \text { Now } \\
& \frac{a_1 e_1}{a_1 e_2}=\frac{1}{3} \Rightarrow a_1 e_2=3 \sqrt{3} \\
& a_1-a_2=2 \\
& a_1 e_2-a_2 e_2=2 e_2 \\
& 3 \sqrt{3}-\sqrt{3}=2 e_2
\end{aligned}
\)

\(
\begin{aligned}
& 2 e_2=2 \sqrt{3} \Rightarrow e_2=\sqrt{3} \Rightarrow a_2=1 \\
& e_1=\frac{1}{\sqrt{3}} \\
& \Rightarrow a_1=3 \\
& b_1^2=a_1^2\left(1-e_1^2\right) \\
& =9\left(1-\frac{1}{3}\right)= 9 \times \frac{2}{3}=6 \\
& b_2^2=a_2^2\left(e_2^2-1\right)=1(3-1)=2
\end{aligned}
\)

\(
\begin{aligned}
&\text { Now Sum }\\
&\begin{aligned}
L R= & \frac{2 b_1^2}{a_1}+\frac{2 b_2^2}{a_2} \\
\end{aligned}
\end{aligned}
\)

$$\begin{gathered} =\frac{2.6}{3}+\frac{2.2}{1} \\ =8 \end{gathered}$$

Given: $2ae=8, e=\frac{4}{5}\Rightarrow a=5$

Since, $b^2=a^2\left(1-e^2\right)$

$$\begin{gathered} \Rightarrow b^2=a^2\times \frac{9}{25} \\ \Rightarrow b^2=9 \end{gathered}$$

$E_1:\frac{x^2}{25}+\frac{y^2}{9}=1$

$l_1:\frac{2b^2}{a}=\frac{2\times 9}{5}=\frac{18}{5}$

Now, $A^2=B^2\left(1-e^2\right)\Rightarrow A^2=\frac{9}{25}{B}^2\Rightarrow A=\frac{3}{5}B$

$2l_1^2=9l_2\Rightarrow 2{\left(\frac{18}{5}\right)}^2=9l_2$

$\Rightarrow l_2=\frac{4\times 18}{25}$

$\Rightarrow \frac{2A^2}{B}=\frac{72}{25}\Rightarrow A^2=\frac{36}{25}B$

$\Rightarrow \frac{9}{25}{B}^2=\frac{36B}{25}\Rightarrow B=4,$

Distance between focii,

\(2Be\) $=2\times \frac{4}{5}\times 4=\frac{32}{5}$

Given Distance between Focl of ellipse \(=2\sqrt {3}\)

\(
\begin{array}{r}
\Rightarrow \quad 2 a_1 e_1=2 \sqrt{3} \\
\Rightarrow a_1 e_1=\sqrt{3}
\end{array}
\)

and for hyperbola \(\Rightarrow \quad 2 a_2 e_2=2 \sqrt{3}\)

\(
\begin{aligned}
& a_2 e_2=\sqrt{3} \\
& \text { and } \frac{e_1}{e_2}=\frac{1}{3} \text { and } a_1-a_2=2 \\
& \text { Now } \\
& \frac{a_1 e_1}{a_1 e_2}=\frac{1}{3} \Rightarrow a_1 e_2=3 \sqrt{3} \\
& a_1-a_2=2 \\
& a_1 e_2-a_2 e_2=2 e_2 \\
& 3 \sqrt{3}-\sqrt{3}=2 e_2
\end{aligned}
\)

\(
\begin{aligned}
& 2 e_2=2 \sqrt{3} \Rightarrow e_2=\sqrt{3} \Rightarrow a_2=1 \\
& e_1=\frac{1}{\sqrt{3}} \\
& \Rightarrow a_1=3 \\
& b_1^2=a_1^2\left(1-e_1^2\right) \\
& =9\left(1-\frac{1}{3}\right)= 9 \times \frac{2}{3}=6 \\
& b_2^2=a_2^2\left(e_2^2-1\right)=1(3-1)=2
\end{aligned}
\)

\(
\begin{aligned}
&\text { Now Sum }\\
&\begin{aligned}
L R= & \frac{2 b_1^2}{a_1}+\frac{2 b_2^2}{a_2} \\
\end{aligned}
\end{aligned}
\)

$$\begin{gathered} =\frac{2.6}{3}+\frac{2.2}{1} \\ =8 \end{gathered}$$