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If $P$ is $(3,1)$ and $Q$ is a point on the curve $y^2=8 x$, then the locus of the mid-point of the line segment $P Q$ is
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The correct answer is:
$4 y^2-16 x-4 y+25=0$
$$
\text {} P(3,1) \text { and } Q \text { is a point of parabola } y^2=8 x
$$
$$
\text { Let } R_{(h, k)} \text { is the mid-point of } P Q \text {. }
$$
So, coordinate of $Q(2 h-3,2 k-1)$ point $Q$ lie on parabola, so $(2 k-1)^2=8(2 h-3)$
$$
\begin{array}{ll}
\Rightarrow & 4 k^2-4 k+1=16 h-24 \\
\Rightarrow & 4 k^2-4 k-16 h+25=0
\end{array}
$$
Hence, locus of $R(h, k)$ is $4 y^2-4 y-16 x+25=0$
\text {} P(3,1) \text { and } Q \text { is a point of parabola } y^2=8 x
$$
$$
\text { Let } R_{(h, k)} \text { is the mid-point of } P Q \text {. }
$$
So, coordinate of $Q(2 h-3,2 k-1)$ point $Q$ lie on parabola, so $(2 k-1)^2=8(2 h-3)$
$$
\begin{array}{ll}
\Rightarrow & 4 k^2-4 k+1=16 h-24 \\
\Rightarrow & 4 k^2-4 k-16 h+25=0
\end{array}
$$
Hence, locus of $R(h, k)$ is $4 y^2-4 y-16 x+25=0$
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