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Q.
If $p$ is the length of perpendicular from the origin on the line $\frac{x}{a}+\frac{y}{b}=1$ and $a^2, p^2, b^2$ are in $AP$, then
Straight Lines
Solution:
Given equation of line is $\frac{x}{a}+\frac{y}{b}-1=0 ....$(i)
At point $(0,0)$, perpendicular distance from line (i) is
$p=\left|\frac{0+0-1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}\right|$
$(\because$ perpendicular distance, $\left.d=\left|\frac{a x_1+b y_1+c}{\sqrt{a^2+b^2}}\right|\right)$
$\Rightarrow p=\frac{1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}} $
$\Rightarrow \frac{1}{p}=\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}$
On squaring both sides, we get
$\Rightarrow \frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2} .....$(ii)
Also, $a^2, p^2, b^2$ are in AP.
$\therefore 2 p^2=a^2+b^2......$(iii)
(using formula, if $a, b$ and $c$ are in $A P$, then $2 b=a+c$ )
Now, from Eq. (ii), we get
$ \frac{1}{p^2}=\frac{a^2+b^2}{a^2 b^2} \Rightarrow p^2=\frac{a^2 b^2}{a^2+b^2} $
$\Rightarrow \frac{a^2+b^2}{2} =\frac{a^2 b^2}{a^2+b^2} $ [using Eq. (iii)]
$\Rightarrow \left(a^2+b^2\right)^2 =2 a^2 b^2$
$\Rightarrow a^4+b^4+2 a^2 b^2 =2 a^2 b^2 \left[\because(a+b)^2=a^2+b^2+2 a b\right] $
$\Rightarrow a^4+b^4 =0$