Question:

Let $P$ be the point $(1, 0)$ and $Q$ a point on the locus $y^2 = 8x$. The locus of mid point of $PQ$ is:

Updated On: Jul 5, 2022
  • $x^2-4y+2 = 0$
  • $x^2+4y+2 = 0$
  • $y^2 + 4x + 2 = 0$
  • $y^2 - 4x + 2 = 0$
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The Correct Option is D

Solution and Explanation

The co-ordinates of $P$ are $(1,0)$. A general point $Q$ on $y^2 = 8x$ is $(2t^2, 4t)$. Mid point of $PQ$ is $(h, k)$ so $2h=2t^{2}+1\,...\left(i\right)$ and $2k=4t \Rightarrow t=k/2\,...\left(ii\right)$ On putting the value of $t$ from E $\left(ii\right)$ in E $\left(i\right)$, we get $2h=\frac{2k^{2}}{4}+1$ $\Rightarrow 4h=k^{2}+2$ So the locus of $\left(h, k\right)$ is $y^{2} - 4x + 2 = 0.$
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