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Let $P$ be the point $(1,0)$ and $Q$ a point on the locus $y^2=8 x$. The locus of mid point of $P Q$ is
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Verified Answer
The correct answer is:
$y^2-4 x+2=0$
$y^2-4 x+2=0$
$\mathrm{P}=(1,0)$
$Q=(h, k)$ such that $k^2=8 h$
Let $(\alpha, \beta)$ be the midpoint of PQ
$$
\begin{aligned}
& \alpha=\frac{\mathrm{h}+1}{2}, \quad \beta=\frac{\mathrm{k}+0}{2} \\
& 2 \alpha-1=\mathrm{h}, \quad 2 \beta=\mathrm{k} . \\
& (2 \beta)^2=8(2 \alpha-1) \Rightarrow \beta^2=4 \alpha-2 \\
& \Rightarrow \mathrm{y}^2-4 \mathrm{x}+2=0 .
\end{aligned}
$$
$Q=(h, k)$ such that $k^2=8 h$
Let $(\alpha, \beta)$ be the midpoint of PQ
$$
\begin{aligned}
& \alpha=\frac{\mathrm{h}+1}{2}, \quad \beta=\frac{\mathrm{k}+0}{2} \\
& 2 \alpha-1=\mathrm{h}, \quad 2 \beta=\mathrm{k} . \\
& (2 \beta)^2=8(2 \alpha-1) \Rightarrow \beta^2=4 \alpha-2 \\
& \Rightarrow \mathrm{y}^2-4 \mathrm{x}+2=0 .
\end{aligned}
$$
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