(1944)

We have, $F_1\equiv (–ae,0)\equiv P(–3,0) \Rightarrow ae=3$

$L_1{L}_2=\frac{2b^2}{a}$

In $△F_1{F}_2{L}_2$,

$\tan 45°=\frac{L_2{F}_2}{F_1{F}_2}=\frac{b^2/a}{2ae}$

$\Rightarrow 2ae=\frac{b^2}{a}$

$\Rightarrow b^2=6a$          [$\because$ ae = 3]

Also, $a^2{e}^2=a^2+b^2$

$\Rightarrow 9=a^2+6a$          [$\because ae=3 \mathrm{and} b^2=6a$]

$\Rightarrow a^2+6a–9=0$

$a=–3\pm 3\sqrt{2}=–3(1\pm \sqrt{2})$

Now, $a^2{b}^2=a^2·6a=6a^3$           [$\because b^2=6a$]

                      $=6(135\sqrt{2}–189)$

On comparing, we get $\alpha$ = 810 and $\beta$ = 1134

$\therefore \alpha +\beta =1944$.

(189)

Given, focus = (–5, 0)

$\Rightarrow ae=5 \mathrm{and} \frac{a}{e}=\frac{9}{5}$

Also, $a=3, e=\frac{5}{3} \mathrm{and} b=4$

$\Rightarrow \frac{x^2}{9}–\frac{y^2}{16}=1$

Since, hyperbola passes through $(\alpha ,2\sqrt{5})$, we get

$\frac{{\alpha }^2}{9}–\frac{4\left(5\right)}{16}=1$

$\Rightarrow {\alpha }^2=9\left(\frac{36}{16}\right)=\frac{81}{4}$

Now, $P=e^2{\alpha }^2–a^2=\frac{25}{9}\times \frac{81}{4}–9=\frac{189}{4}$

$\therefore 4p=189$.

(19)

Since, centre of circle lies on positive x-axis and one of the foci of hyperbola $\frac{x^2}{{\alpha }^2}–\frac{y^2}{{\beta }^2}=1$ are same.

$\therefore$   Centre of circle = $(\alpha ,0)$

Since, x – y + 1 = 0 is tangent to the circle.

$\therefore$$\left|\frac{\alpha +1}{\sqrt{2}}\right|$$=r$          [where 'r' is the radius of circle]

$\Rightarrow {(\alpha +1)}^2=2r^2$          ... (i)

Also, –3x + 2y = 1 is the chord of the circle

$\therefore$$|\frac{–3\alpha –1}{\sqrt{9+4}}{|}^2$${+\left(\frac{2}{\sqrt{13}}\right)}^2=r^2$          $[\because \mathrm{Length} \mathrm{of} \mathrm{chord} = \frac{4}{\sqrt{13}}]$

$\therefore \frac{{(3\alpha +1)}^2}{13}+\frac{4}{13}=r^2$

$\Rightarrow \frac{9{\alpha }^2+1+6\alpha +4}{13}=\frac{({\alpha }^2+1+2\alpha )}{2}$

$\Rightarrow 5{\alpha }^2–14\alpha –3=0$          

$\Rightarrow \alpha =3$                                        [$\because \alpha >0$]

$\therefore r=2\sqrt{2}$

Now, $\alpha e=3 \mathrm{and} 2\alpha =4\sqrt{2}$

$\therefore {\alpha }^2{e}^2=9 \mathrm{and} \alpha =2\sqrt{2}$

$\Rightarrow {\alpha }^2(1+\frac{{\beta }^2}{{\alpha }^2})=9 \Rightarrow 8+{\beta }^2=9 \Rightarrow {\beta }^2=1$

$\therefore 2{\alpha }^2+3{\beta }^2=2\left(8\right)+3\times 1=16+3=19$.

(55)

Given, Hyperbola : $H_1:\frac{x^2}{a^2}–\frac{y^2}{b^2}=1$ and $e_1=\sqrt{\frac{5}{2}}$ and $H_2:\frac{–x^2}{A^2}+\frac{y^2}{B^2}=1$

Using $H_1$, length of latus rectum = $15\sqrt{2}$

$\Rightarrow \frac{2b^2}{a}=15\sqrt{2}$          ... (i)

Since, $e_1=\sqrt{\frac{5}{2}} \Rightarrow 1+\frac{b^2}{a^2}=\frac{5}{2} \Rightarrow \frac{b^2}{a^2}=\frac{3}{2}$           ... (ii)

Using (i) and (ii), we get $a=5\sqrt{2} \mathrm{and} b=5\sqrt{3}$

Now, for $H_2, \frac{2A^2}{B}=12\sqrt{5}$          ... (iii)

Since, product of transverse axes is $100\sqrt{10}$, then

      $2a·2B=100\sqrt{10}$

$\Rightarrow 2\times 5\sqrt{2}\times 2B=100\sqrt{10} \Rightarrow B=5\sqrt{5}$

$\therefore A=5\sqrt{6}$          [Using (iii)]

Now, eccentricity of $H_2$ is given by

$e_2^2=1+\frac{A^2}{B^2}=1+\frac{150}{125}=1+\frac{30}{25}=\frac{55}{25}$

$\Rightarrow 25e_2^2=55$.