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ABC is a right angled triangle , with C = $90^{\circ}$ . If P is the perpendicular from C to AB and a,b,c have the usual meaning, then find the value of $\frac{1}{a^{2}} + \frac{1}{b^{2}}$ .

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p^{2}

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

p

\frac{1}{p}

answer is B.

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Detailed Solution

Given ABC is a right-angled triangle with the right angle at C.
Let P is the perpendicular from C to AB and it intersects AB at D . Given, the length of the side AB = c, BC = a, CA = b and CD = p .
Area of triangle ABC =

\frac{1}{2} \times BC \times AC = \frac{1}{2} ab

Also, Area of triangle ABC =

\frac{1}{2} \times AB \times CD = \frac{1}{2} cp

\Rightarrow ab = cp

Since

a^{2} + b^{2} = c^{2} = {(\frac{ab}{p})}^{2} = \frac{a^{2} b^{2}}{p^{2}}

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

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