Question

If 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, then the locus of the foot of perpendicular from the origin on the straight line is

• A. straight line
• B. parabola
• C. ellipse
• D. circle

Answer 1

Answer: (D)

Variable line is
$\frac{x}{a}+\frac{y}{b}=1$ .... (1)
Any line perpendicular to (1) and passing through the origin will be
$\frac{x}{b}-\frac{y}{a}=0$ ..... (2)
Now foot of the perpendicular from the origin to line (1) is the point of intersection (1) and (2).
Let it be $P(\alpha ,\beta )$, then
$\frac{\alpha }{a}+\frac{\beta }{b}=1$ ..... (3)
and $\frac{\alpha }{b}-\frac{\beta }{a}=0$ ......(4)
Squaring and adding $(3)$ and (4), we get
${\alpha }^2\left(\frac{1}{a^2}+\frac{1}{b^2}\right)+{\beta }^2\left(\frac{1}{b^2}+\frac{1}{a^2}\right)=1$
$\therefore \left({\alpha }^2+{\beta }^2\right)\frac{1}{c^2}=1.$
Hence, the locus of $P(\alpha ,\beta )$ is $x^2+y^2=c^2$.