P is a point on $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ and A, A’ are the vertices of the conic. If PA, PA’ meet an asymptote at K and L then $(K L)^{2} =$

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Solution

The correct option is D
P is a point on the hyperbola $\Rightarrow P = (a s e c θ, b t a n θ)$
A, A’ are the vertices of the hyperbola $\Rightarrow A (a, 0), A^{'} (- a, 0)$
Equation of PA is $y = \frac{b t a n θ}{a (s e c θ - 1)} (x - a) \to (1)$
Equation of PA’ is y $= \frac{b t a n θ}{a (s e c θ - 1)} (x + a) \to (2)$
Equation of an asymptote is $\frac{x}{a} - \frac{y}{b} = 0 \Rightarrow y = \frac{b}{a} x \to (3)$
Now K is the point of intersection of (1) and (3) and L is the point of intersection of (2) and (3).
$∴ K = (\frac{- a t a n θ}{s e c θ - t a n θ - 1}, \frac{- b t a n θ}{s e c θ - t a n θ + 1}), L = (\frac{a t a n θ}{s e c θ - t a n θ + 1}, \frac{b t a n θ}{s e c θ - t a n θ + 1})$
$(K L)^{2} = {(\frac{a t a n θ}{s e c θ - t a n θ + 1} + \frac{a t a n θ}{s e c θ - t a n θ - 1})}^{2} + (\frac{b t a n θ}{s e c θ - t a n θ + 1} + \frac{b t a n θ}{s e c θ - t a n θ - 1})$
$= (a^{2} t a n^{2} θ + b^{2} t a n^{2} θ) {(\frac{1}{s e c θ - t a n θ + 1} + \frac{1}{s e c θ - t a n θ - 1})}^{2}$
$= (a^{2} + b^{2}) t a n^{2} θ {[\frac{(s e c θ - t a n θ)^{2} - 1 + s e c θ - t a n θ + 1}{(s e c θ - t a n θ)^{2} - 1}]}^{2}$
$= (a^{2} + b^{2}) t a n^{2} θ {[\frac{2 (s e c θ - t a n θ)}{s e c^{2} θ + t a n^{2} θ - 2 s e c θ t a n θ - 1}]}^{2}$
$= (a^{2} + b^{2}) t a n^{2} θ {[\frac{2 (s e c θ - t a n θ)}{- 2 t a n θ (s e c θ - t a n θ)}]}^{2} = a^{2} + b^{2}$

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