Let $d_{1}$ and $d_{2}$ be the lengths of perpendiculars drawn from foci $S$ and $S$ ' of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ to the tangent at any point P on the ellipse. Then, $S P : S P' =$

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$d_{1}^{2}$

none of these

$d_{2} : d_{1}$

$d_{1} : d_{2}$

answer is A.

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Detailed Solution

Let $\frac{x}{a} \cos θ + \frac{y}{b} \sin θ = 1$ be a tangent to the ellipse

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ at point $P (a \cos θ, b \sin θ)$ Then

$d_{1} =∣ \frac{e \cos θ - 1}{\sqrt{\frac{\cos^{2} θ}{a^{2}} + \frac{\sin^{2} θ}{b^{2}}}} and d_{2} = |\frac{e \cos θ + 1}{\sqrt{\frac{\cos^{2} θ}{a^{2}} + \frac{\sin^{2} θ}{b^{2}}}}|$

$\begin{array}{ll} \Rightarrow d_{1} : d_{2} = | e \cos θ - 1 | : | e \cos θ + 1 | \\ \Rightarrow d_{1} : d_{2} = | a e \cos θ - a | : | a e \cos θ + a | \\ \Rightarrow d_{1} : d_{2} = S P : S^{'} P . \end{array}$