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(B) Let $R(h, k)$ be the mid-point of $OM$. Let the coordinates of $M$ be $(\alpha, \beta)$.Since $R$ is the mid-point of $OM$,we have $\left(\frac{0+

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$2x^2+2y^2-ax-by=0$

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(B) Let $R(h, k)$ be the mid-point of $OM$. Let the coordinates of $M$ be $(\alpha, \beta)$.Since $R$ is the mid-point of $OM$,we have $\left(\frac{0+\alpha}{2}, \frac{0+\beta}{2}ight) = (h, k)$,which implies $\alpha = 2h$ and $\beta = 2k$.Thus,the coordinates of $M$ are $(2h, 2k)$.The slope of $OM$ is $m_1 = \frac{2k-0}{2h-0} = \frac{k}{h}$.The slope of the line $MQ$ (which is the line $L$) is $m_2 = \frac{2k-b}{2h-a}$.Since $OM \perp MQ$,the product of their slopes is $-1$:$\frac{k}{h} 	imes \frac{2k-b}{2h-a} = -1$$k(2k-b) = -h(2h-a)$$2k^2 - bk = -2h^2 + ah$$2h^2 + 2k^2 - ah - bk = 0$Replacing $(h, k)$ with $(x, y)$,the locus is $2x^2 + 2y^2 - ax - by = 0$.

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

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