Question

If $O$ is the origin and $A$ is a point on the curve $y^2=4x$. Then, the locus of the mid - point of $OA$, is

• A. $x^2=4y$
• B. $x^2=2y$
• C. $x^2=16y$
• D. $y^2=2x$

Answer 1

The correct option is D

$y^2=2x$

Step 1: Apply the mid-point formula

According to the mid-point formula,

$\mathit{x}\mathbf{=}\frac{{\mathbf{x}}_{\mathbf{1}}\mathbf{+}{\mathbf{x}}_{\mathbf{2}}}{\mathbf{2}}\mathbf{,}\mathit{y}\mathbf{=}\frac{{\mathbf{y}}_{\mathbf{1}}\mathbf{+}{\mathbf{y}}_{\mathbf{2}}}{\mathbf{2}}$, where $x\mathrm{and}y$ is the midpoint of the endpoints $x_1\text{and}{x}_2,y_1\mathrm{and}{y}_2$ respectively.

We know that the coordinate of the origin is $\left(0,0\right)$ and the coordinate of the point $A$ will be $\left(x,4\right)$

Then, suppose the two coordinates are $\left(x_1,y_1\right)\text{and}\left(x_2,y_2\right)$.

Step 2: Calculation of the locus of the mid-point

By the midpoint formula, we have

$\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$

From this, we can write that,

$x_1=\frac{x}{2}\text{and}{y}_1=\frac{y}{2}$

Therefore,

$x=2x_1\text{and}y=2y_1$

Now, as per the given equation $y^2=4x$

$$\begin{gathered} \Rightarrow {\left(2y_1\right)}^2=4\left(2x_1\right) \\ \Rightarrow 4{y_1}^2=8x_1 \\ \Rightarrow {y_1}^2=2x_1 \end{gathered}$$

$\therefore y^2=2x$

Hence, the correct option is (D).