Question

If the length of perpendicular drawn from origin to $\frac{x}{2a}+\frac{y}{2b}=4$ is $p$ then prove that $\frac{64}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}$.

Answer 1

$\frac{x}{2a}+\frac{y}{2b}=4$

Length of perpendicular drawn from origin to the line $\frac{x}{2a}+\frac{y}{2b}=4$ is

$p=|\frac{0+0-4}{\sqrt{\frac{1}{4a^2}+\frac{1}{4b^2}}}|$

$⟹\frac{8}{p}=\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}$

$⟹\frac{64}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}$

Hence proved.