Question

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

• A. $a^2+b^2=0$
• B. $a^3+b^3=0$
• C. $a^4+b^4=0$
• D. $a^5+b^5=0$

Answer 1

Answer: (C)

Given equation of line is $\frac{x}{a}+\frac{y}{b}-1=\mathrm{0....}$(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, $d=\left|\frac{a{x}_1+b{y}_1+c}{\sqrt{a^2+b^2}}\right|)$
$\Rightarrow p=\frac{1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}$
$\Rightarrow \frac{1}{p}=\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}$
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.
$\therefore 2p^2=a^2+b^2......$(iii)
(using formula, if $a,b$ and $c$ are in $AP$, then $2b=a+c$ )
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)]
$\Rightarrow {\left(a^2+b^2\right)}^2=2a^2{b}^2$
$\Rightarrow a^4+b^4+2a^2{b}^2=2a^2{b}^2\left[\because (a+b{)}^2=a^2+b^2+2ab\right]$
$\Rightarrow a^4+b^4=0$