Hyperbola

Equation of tangent to ellipse ${{{x^2}} \over 4} + {{{y^2}} \over 9} = 1$ and given slope m is : $y = mx + \sqrt {4{m^2} + 9} $ ..... (i)

For slope m equation of tangent to hyperbola is :

$y = mx + \sqrt {42{m^2} - 143} $ ....... (ii)

Tangents from (i) and (ii) are identical then

$4{m^2} + 9 = 42{m^2} - 143$

$\therefore$ $m = \, \pm \,2$ (+2 is not acceptable)

$\therefore$ $m = - 2$.

Hence, ${x_1} = {8 \over 5}$ and ${x_2} = {{84} \over 5}$

$\therefore$ $|2{x_1} + {x_2}| = \left| {{{16} \over 5} + {{84} \over 5}} \right| = 20$

Vertices of hyperbola $ = (0,\, \pm \,8)$

As ellipse pass through it i.e.,

$0 + {{64} \over {{b^2}}} = 1 \Rightarrow {b^2} = 64$ ...... (1)

As major axis of ellipse coincide with transverse axis of hyperbola we have b > a i.e.

${e_E} = \sqrt {1 - {{{a^2}} \over {64}}} = {{\sqrt {64 - {a^2}} } \over 8}$

and ${e_H} = \sqrt {1 + {{49} \over {64}}} = {{\sqrt {113} } \over 8}$

$\therefore$ ${e_E}\,.\,{e_H} = {1 \over 2} = {{\sqrt {64 - {a^2}} \sqrt {113} } \over {64}}$

$ \Rightarrow (64 - {a^2})(113) = {32^2}$

$ \Rightarrow {a^2} = 64 - {{1024} \over {113}}$

L.R of ellipse $ = {{2{a^2}} \over b} = {2 \over 8}\left( {{{113 \times 64 - 1024} \over {113}}} \right)$

$ = l = {{1552} \over {113}}$

$\therefore$ $113l = 1552$

$2a + 2b = 4\left( {2\sqrt 2 + \sqrt {14} } \right)$ ...... (1)

$1 + {{{b^2}} \over {{a^2}}} = {{11} \over {14}}$ ....... (2)

$ \Rightarrow {{{b^2}} \over {{a^2}}} = {7 \over 4}$ ....... (3)

and $a + b = 4\sqrt 2 + 2\sqrt {14} $ ...... (4)

By (3) and (4)

$ \Rightarrow a + {{\sqrt 7 } \over 2}a = 4\sqrt 2 + 2\sqrt {14} $

$ \Rightarrow {{a\left( {2 + \sqrt 7 } \right)} \over 2} = 2\sqrt 2 \left( {2 + \sqrt 7 } \right)$

$ \Rightarrow a = 4\sqrt 2 \Rightarrow {a^2} = 32$ and ${b^2} = 56$

$ \Rightarrow {a^2} + {b^2} = 32 + 56 = 88$

Equation of L₁ is

${{x\sec \theta } \over 4} - {{y\tan \theta } \over 2} = 1$ ..... (i)

Equation of line L₂ is

${{x\tan \theta } \over 2} + {{y\sec \theta } \over 4} = 0$ ..... (ii)

$\because$ Required point of intersection of L₁ and L₂ is (x₁, y₁) then

${{{x_1}\sec \theta } \over 4} - {{{y_1}\tan \theta } \over 2} - 1 = 0$ ...... (iii)

and ${{{y_1}\sec \theta } \over 4} - {{{x_1}\tan \theta } \over 2} = 0$ ...... (iv)

From equations (iii) and (iv)

$\sec \theta = {{4{x_1}} \over {x_1^2 + y_1^2}}$ and $\tan \theta = {{ - 2{y_1}} \over {x_1^2 + y_1^2}}$

$\therefore$ Required locus of (x₁, y₁) is

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

$\therefore$ $\alpha$ = 16, $\beta$ = $-$4

$\therefore$ $\alpha$ + $\beta$ = 12

${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1\left( {e = {5 \over 4}} \right)$

So, ${b^2} = {a^2}\left( {{{25} \over {16}} - 1} \right) \Rightarrow b = {3 \over 4}a$

Also $\left( {{8 \over {\sqrt 5 }},{{12} \over 5}} \right)$ lies on the given hyperbola

So, ${{64} \over {5{a^2}}} - {{144} \over {25\left( {{{9{a^2}} \over {16}}} \right)}} = 1 \Rightarrow a = {8 \over 5}$ and $b = {6 \over 5}$

Equation of normal

${{64} \over {25}}\left( {{x \over {{8 \over {\sqrt 5 }}}}} \right) + {{36} \over {25}}\left( {{y \over {{{12} \over 5}}}} \right) = 4$

$ \Rightarrow {8 \over {5\sqrt 5 }}x + {3 \over 5}y = 4$

$ \Rightarrow 8\sqrt 5 x + 15y = 100$

So, $\beta$ = 15 and $\lambda$ = 100

Gives $\lambda$ $-$ $\beta$ = 85

$\because$ $H:{{{x^2}} \over {{a^2}}} - {{{y^2}} \over 1} = 1$

$\therefore$ Length of latus rectum $ = {2 \over a}$

$E:{{{x^2}} \over 4} + {{{y^2}} \over 3} = 1$

Length of latus rectum $ = {6 \over 2} = 3$

$\because$ ${2 \over a} = 3 \Rightarrow a = {2 \over 3}$

$\therefore$ $12\left( {e_H^2 + e_E^2} \right) = 12\left( {1 + {9 \over 4}} \right) + \left( {1 - {3 \over 4}} \right) = 42$