Question

The line $\frac{x}{a}+\frac{y}{b}=1$ moves in such a way that $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}$ where $c$ is a constant. The locus of foot of perpendicular from the origin on the given line will be

• A. $x^2+y^2=a^2$
• B. $x^2+y^2=b^2$
• C. $x^2+y^2=c^2$
• D. $x^2+y^2=(a^2-b^2{)}^2$

Answer 1

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

Let locus of foot of perpendicular from $(0,0)$ upon line $bx+ay-ab=0$ be$(h,k)$
$\Rightarrow \frac{h-0}{b}=\frac{k-0}{a}=-\frac{-ab}{a^2+b^2}\Rightarrow h=\frac{a{b}^2}{a^2+b^2},k=\frac{a^{2}b}{a^2+b^2}\Rightarrow h^2+k^2=\frac{a^2{b}^2}{a^2+b^2}=c^2$

$\therefore$ required locus equation is $x^2+y^2=c^2$