If $' p'$ is the perpendicular distance from the origin to the straight line $\frac{x}{a} + \frac{y}{b} = 1$ , then $\frac{1}{a^{2}} + \frac{1}{b^{2}}$ is

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\frac{2}{p}

\frac{1}{p^{2}}

\frac{1}{p}

\frac{2}{p^{2}}

answer is B.

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Detailed Solution

As per the problem

$\frac{1}{2} \times a \times b = \frac{1}{2} \times \sqrt{a^{2} + b^{2}} \times p$

$\frac{a^{2} b^{2}}{a^{2} + b^{2}} = p^{2}$

$\frac{a^{2}}{a^{2} b^{2}} + \frac{b^{2}}{a^{2} b^{2}} = \frac{1}{p^{2}}$

$\frac{1}{b^{2}} + \frac{1}{a^{2}} = \frac{1}{p^{2}}$

(or)

perpendicular distance from the origin= $\frac{| c |}{\sqrt{a^{2} + b^{2}}}$

$p = \frac{| - 1 |}{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}}}}$

$\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}}} = \frac{1}{p}$

$\frac{1}{a^{2}} + \frac{1}{b^{2}} = \frac{1}{p^{2}}$

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