Question

If $M$ is the foot of the perpendicular drawn from the origin $O$ on to the variable line L,passing through a fixed point $(a,b)$ then the locus of the mid point of $OM$ is

• A. $x^2+y^2=a^2+b^2$
• B. $2x^2+2y^2-ax-by=0$
• C. $ax+by=0$
• D. $2x^2+2y^2-ay-bx=0$

Answer 1

Answer: (B)

We have, $R(h,k)$ is mid-point of $OM$.
$\therefore \left(\frac{0+\alpha }{2},\frac{0+\beta }{2}\right)=(h,k)$
$\Rightarrow \alpha =2h,\beta =2k$
$\Rightarrow$ Coordinates of $M$ are $(2h,2k)$.
Now, slope of $OM=\frac{2k-0}{2h-0}=\frac{k}{h}$
and slope of $MQ=\frac{2k-b}{2h-a}$
Since, $OM\perp MQ$
$\therefore \frac{k}{h}\times \frac{2k-b}{2h-a}=-1$
$\Rightarrow 2k^2-bk=-2h^2+ah$
$\Rightarrow 2h^2+2k^2-ah-bk=0$
$\therefore$ Locus of $R(h,k)$ is
$2x^2+2y^2-ax-by=0$