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Let $P$ be the mid-point of a chord joining the vertex of the parabola $y^{2}=8 x$ to another point on it. Then, the locus of $P$ is
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The correct answer is:
$y^{2}=4 x$
Let the chord be MN. Let other end of chord joining from vertex to it lying on $y^{2}=8 x$ be $N\left(2 t^{2}, 4 t\right)$
$\therefore$ Mid-point of $M N=\left(t^{2}, 2 t\right)$ $\therefore \quad x=t^{2}, y=2 t \Rightarrow x=\left(\frac{y}{2}\right)^{2}$
$\Rightarrow \quad y^{2}=4 x$
is the required locus of $P$
$\therefore$ Mid-point of $M N=\left(t^{2}, 2 t\right)$ $\therefore \quad x=t^{2}, y=2 t \Rightarrow x=\left(\frac{y}{2}\right)^{2}$
$\Rightarrow \quad y^{2}=4 x$
is the required locus of $P$
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