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Question Number 29960 by rahul 19 last updated on 14/Feb/18

	Answered by ajfour last updated on 15/Feb/18

	$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \Rightarrow \frac{x}{a^{2}} + \frac{y}{b^{2}} (\frac{d y}{d x}) = 0$ $s l o p e o f a n o r m a l = - \frac{d x}{d y} = n$ $n = \frac{a^{2}}{b^{2}} (\frac{y}{x}) a n d i f w e l e t$ $x = a \cos θ, y = b \sin θ t h e n$ $n = \frac{a \sin θ}{b \cos θ}$ $e q . o f a n o r m a l t h r o u g h (h, k) i s$ $y - k = \frac{a \sin θ}{b \cos θ} (x - h)$ $\Rightarrow (b \sin θ - k) b \cos θ = a \sin θ (a \cos θ - h)$ $- k b \cos θ = \sin θ (a^{2} \cos θ - b^{2} \cos θ - a h)$ $k^{2} b^{2} \cos^{2} θ = (1 - \cos^{2} θ) {[(a^{2} - b^{2}) \cos θ - a h]}^{2}$ $l e t u s s a y \cos θ = t$ $k^{2} b^{2} t^{2} = (1 - t^{2}) [{(a^{2} - b^{2})}^{2} t^{2} + a^{2} h^{2}$ $- 2 a h (a^{2} - b^{2}) t]$ $k^{2} b^{2} t^{2} = {(a^{2} - b^{2})}^{2} t^{2} + a^{2} h^{2}$ $- 2 a h (a^{2} - b^{2}) t - {(a^{2} - b^{2})}^{2} t^{4}$ $- a^{2} h^{2} t^{2} + 2 a h (a^{2} - b^{2}) t^{3}$ $f i n a l l y$ ${(a^{2} - b^{2})}^{2} t^{4} - 2 a h (a^{2} - b^{2}) t^{3}$ $+ {k t}^{2} + 2 a h (a^{2} - b^{2}) t - a^{2} h^{2} = 0$ $Σ \cos α = Σ t_{1} = \frac{2 a h}{(a^{2} - b^{2})}$ $Σ \sec α = \frac{Σ \cos α \cos β \cos γ}{Π \cos α}$ $= \frac{Σ t_{1} t_{2} t_{3}}{Π t_{1}} = \frac{(\frac{2 a h}{a^{2} - b^{2}})}{[\frac{a^{2} h^{2}}{{(a^{2} - b^{2})}^{2}}]} = \frac{2 (a^{2} - b^{2})}{a h}$ $\Rightarrow (Σ \cos α) (Σ \sec α) =$ $\frac{2 a h}{(a^{2} - b^{2})} \times \frac{2 (a^{2} - b^{2})}{a h} = 4 .$
	Commented by rahul 19 last updated on 15/Feb/18

	$thank u sir!$
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