If $ϕ$ is the angle between the diameter through any point on a standard ellipse and the normal at that point, then the greatest value of tan $ϕ$ is

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$\frac{a^{2} - b^{2}}{2 ab}$

$\frac{b^{2}}{a^{2}}$

$\frac{2 ab}{a^{2} + b^{2}}$

$\frac{a^{2} + b^{2}}{ab}$

answer is C.

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Detailed Solution

$L e t e l l i p s e b e \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 L e t P = (a \cos θ, b \sin θ) b e a n y p o i n t o n e l l i p s e N o w e q u a t i o n o f t h e d i a m e t r e C P i s y = (\frac{b}{a} \tan θ) x s l o p e (m_{1}) = \frac{b}{a} \tan θ N o w e q u a t i o n o f n o r m a l a t P i s a x s e c θ - b y \cos e c θ = a^{2} e^{2} s l o p e (m_{2}) = \frac{a}{b} \tan θ N o w \tan φ = \frac{|\frac{b}{a} \tan θ - \frac{a}{b} \tan θ|}{1 + (\frac{b}{a} \tan θ) (\frac{a}{b} \tan θ)} = \frac{\frac{a^{2} - b^{2}}{a b} \tan θ}{s e c^{2} θ} = \frac{a^{2} - b^{2}}{2 a b} (2 \sin θ \cos θ) = \frac{a^{2} - b^{2}}{2 a b} (\sin 2 θ) N o w g r e a t v a l u e o f \tan φ = \frac{a^{2} - b^{2}}{2 a b} (1) = \frac{a^{2} - b^{2}}{2 a b}$