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 $x^{2}+y^{2}=c^{2}$.