Question

If $\frac{x}{a}+\frac{y}{b}=1$ is a variable line where $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}$ ( $c$-constant) , then locus of the foot of the perpendicular drawn from origin is

• A. $x^2+y^2=c^2$
• B. $x^2+y^2=2c^2$
• C. $x^2+y^2=2$
• D. $x^2+y^2=\frac{1}{c^2}$

Answer 1

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

$\frac{x}{a}+\frac{y}{b}=1$ and $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}$
Foot of perpendicular from $(0,0)$ to $\frac{x}{a}+\frac{y}{b}=1$ is $P(h,k)$
So,
$\frac{h-0}{\frac{1}{a}}=\frac{k-0}{\frac{1}{b}}=-\frac{(\frac{0}{a}+\frac{0}{b}-1)}{(\frac{1}{a^2}+\frac{1}{b^2})}=c^2$
$\therefore h=\frac{c^2}{a}$ ----(1)
$k=\frac{c^2}{b}$ ----(2)
$h^2+k^2=c^4(\frac{1}{a^2}+\frac{1}{b^2})=\frac{c^4}{c^2}=c^2$