Question

For a variable line $\frac{x}{a}+\frac{y}{b}=1$ where $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}$, the locus of the foot of perpendicular drawn from origin to it is:

• A. $x^2+y^2=\frac{c^2}{2}$
• B. $x^2+y^2=c^2$
• C. $x^2+y^2=2c$
• D. None of these

Answer 1

The correct option is B $x^2+y^2=c^2$

Let foot of perpendicular be $P(h,k)$

Distance of $\sqrt{h^2+b^2}=\left|\frac{0-1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}\right|$

Squaring,

$h^2+b^2=\frac{1}{\frac{1}{c^2}}$

$\Rightarrow h^2+k^2=c^2$

$\Rightarrow x^2+y^2=c^2$ is locus

$\Rightarrow \left(B\right)$.