If p is the length of perpendicular from the origin

Given equation is

$\frac{x}{a}+\frac{y}{b}=1$

Since, $p$ is the length of perpendicular drawn from the origin to the given line

$\therefore \mathrm{p}=\left|\frac{\frac{0}{\mathrm{a}}+\frac{0}{\mathrm{~b}}-1}{\sqrt{\frac{1}{\mathrm{a}^{2}}+\frac{1}{\mathrm{~b}^{2}}}}\right|$

Squaring both the sides, we have

$\mathrm{p}^{2}=\left|\frac{1}{\frac{1}{\mathrm{a}^{2}}+\frac{1}{\mathrm{~b}^{2}}}\right|$

$\Rightarrow \frac{1}{\mathrm{p}^{2}}=\frac{1}{\mathrm{a}^{2}}+\frac{1}{\mathrm{~b}^{2}}$ $\ldots$ (i)

Since, $a^{2}, b^{2}$ and $p^{2}$ are in $A P$

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

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

$\Rightarrow \frac{1}{\mathrm{p}^{2}}=\frac{2}{\mathrm{a}^{2}+\mathrm{b}^{2}}$ ....(ii)

Form equation (i) and (ii), we get

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

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

⇒ (a2 + b2) (a2 + b2) = 2(a2b2)

⇒ a4 + b4 + a2b2 + a2b2 = 2a2b2

⇒ a4 + b4 = 0

Hence Proved

Objective Type Questions