If $x \cos α + y \sin α = p$ touches $x^{2} + a^{2} y^{2} = a^{2}$ , then

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$p^{2} = a^{2} \cos^{2} α + \sin^{2} α$

$1 / p^{2} = \cos^{2} α + a^{2} \sin^{2} α$

$1 / p^{2} = \sin^{2} α + α^{2} \cos^{2} α$

$p^{2} = a^{2} \sin^{2} α + \cos^{2} α$

answer is B.

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Detailed Solution

Solving for y, we have $y = (p / \sin α) - x \cot α .$ .

Putting this value in $x^{2} + a^{2} y^{2} = a^{2},$ we have

$x^{2} + a^{2} {(\frac{p}{\sin α} - x \cot α)}^{2} = a^{2} \Rightarrow x^{2} (1 + a^{2} \cot^{2} α) - \frac{2 a^{2} x p \cot α}{\sin α} + \frac{a^{2} p^{2}}{\sin^{2} α} - a^{2} = 0$

The discriminant of this equation must be zero. So

$a^{4} \frac{p^{2} \cot^{2} α}{\sin^{2} α} = (1 + a^{2} \cot^{2} α) (\frac{a^{2} p^{2}}{\sin^{2} α} - a^{2}) \begin{array}{l} \Rightarrow a^{2} p^{2} \cot^{2} α = (1 + a^{2} \cot^{2} α) (p^{2} - \sin^{2} α) \\ \Rightarrow p^{2} (a^{2} \cot^{2} α - 1 - a^{2} \cot^{2} α) = - \sin^{2} α - a^{2} \cos^{2} α \\ \Rightarrow p^{2} = a^{2} \cos^{2} α + \sin^{2} α \end{array}$