The product of the focal distances of a point on the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 (a > b)$ from which two chords are drawn which are perpendicular to each other and through the centre is

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$a^{2} (1 - e^{2} \cos^{2} θ)$

$a^{2} \cos^{2} θ + b^{2} \sin^{2} θ$

$a^{2} \sin^{2} θ + b^{2} \cos^{2} θ$

$b^{2} (1 - e^{2} \cos^{2} θ)$

answer is A, B.

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Detailed Solution

$\begin{array}{l} N o w SP . S^{'} P = | (a - aecos θ) (a + aecos θ) | \\ = a^{2} - a^{2} e^{2} \cos^{2} θ \\ = a^{2} (1 - e^{2} \cos^{2} θ) \\ a n d a^{2} - a^{2} e^{2} \cos^{2} θ = a^{2} - (a^{2} - b^{2}) \cos^{2} θ \\ = a^{2} \sin^{2} θ + b^{2} \cos^{2} θ \end{array}$