Question

If $P$ be the point $\left(1,0\right)$ and $Q$ a point on the locus of $y^2=8x$. The locus of mid-point of $PQ$ is

• A. $x^2–4y+2=0$
• B. $x^2+4y+2=0$
• C. $y^2–4x+2=0$
• D. $y^2+4x-2=0$

Answer 1

The correct option is C

$y^2–4x+2=0$

Explanation for the correct option

Step 1: Diagram for the solution

Suppose that the coordinates of the point $Q$ are in the parametric form that are $\left(a{t}^2,2at\right)$. Let the coordinates of the mid-point of $PQ$ be $R$ with coordinates $\left(h,k\right)$.

Step 2: Calculation for the locus of the midpoint of $\mathit{P}\mathit{Q}$

As per the equation of the parabola $y^2=4ax$.

The equation given here is $y^2=8x$,

$\begin{array}{rcl}\therefore 4ax& =& 8x\\ & \Rightarrow & a=2\end{array}$

So the coordinates of the point $Q$ are $\left(2t^2,4t\right)$.

Now, by the mid-point formula, we have

$\begin{array}{rcl}\therefore \left(h,k\right)& =& \left(\frac{1+2t^2}{2},\frac{4t}{2}\right)\\ & =& \left(\frac{1+2t^2}{2},2t\right)\end{array}$

From this, we found that,

$h=\begin{array}{rcl}& & \frac{1+2t^2}{2}\end{array}\text{and}k=2t$

Then $t$ is equal to $\frac{k}{2}$.

$\begin{array}{rcl}\therefore h& =& \frac{1+2t^2}{2}\\ & \Rightarrow & 2h=1+2{\left(\frac{k}{2}\right)}^2\\ & \Rightarrow & 2h=\frac{2+k^2}{2}\\ & \Rightarrow & 4h=2+k^2\\ & \Rightarrow & k^2-4h+2=0\end{array}$

Therefore, the required equation is $y^2–4x+2=0$.

Hence, the correct option is (C)