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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 $O M$ is
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The correct answer is:
$2 x^2+2 y^2-a x-b y=0$
We have, $R(h, k)$ is mid-point of $O M$.
$\therefore \quad\left(\frac{0+\alpha}{2}, \frac{0+\beta}{2}\right)=(h, k) \Rightarrow \alpha=2 h, \beta=2 k$
$\Rightarrow$ Coordinates of $M$ are $(2 h, 2 k)$.
Now, slope of $O M=\frac{2 k-0}{2 h-0}=\frac{k}{h}$ and slope of $M Q=\frac{2 k-b}{2 h-a}$
Since, $O M \perp M Q$
$\begin{aligned}
& \therefore \quad \frac{k}{h} \times \frac{2 k-b}{2 h-a}=-1 \Rightarrow 2 k^2-b k=-2 h^2+a h \\
& \Rightarrow \quad 2 h^2+2 k^2-a h-b k=0 \\
& \therefore \text { Locus of } R(h, k) \text { is } 2 x^2+2 y^2-a x-b y=0
\end{aligned}$
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