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

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Solution

Let the point $C$ be at origin.
Then take $C B$ as $x$ -axis and $C A$ as $y$ -axis.
Then co-ordinate of $C (0, 0), A (0, b)$ and $B (a, 0)$
Equation of line $A B$
$y - 0 = \frac{b - 0}{0 - a} (x - a)$
$- a y = b x - a b$
$b x + a y - a b = 0$
$p =$ perpendicular length at point $C$ on line $A B$
$p = ∣ ∣ ∣ \frac{b \times 0 + a \times 0 - a b}{\sqrt{a^{2} + b^{2}}} ∣ ∣ ∣$
$p = ∣ ∣ ∣ \frac{- a b}{\sqrt{a^{2} + b^{2}}} ∣ ∣ ∣$
Squaring on both side, we get
$p^{2} = \frac{a^{2} b^{2}}{a^{2} + b^{2}}$
$p^{2} a^{2} + p^{2} b^{2} = a^{2} b^{2}$
Dividing both side by $p^{2} a^{2} b^{2}$ , we get
$\frac{1}{b^{2}} + \frac{1}{a^{2}} = \frac{1}{p^{2}}$

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1p2=1a2+1b2

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