If the locus of a point which divides a chord with slope 2 of the parabola $y^2=4 x$, internally in the ratio $1: 3$ is a parabola, then its vertex is

• A

  $(2,1)$
• B

  $\left(\frac{3}{16}, \frac{3}{2}\right)$
• C

  $\left(\frac{3}{4}, \frac{3}{16}\right)$
• Correct answer
  D

  $\left(\frac{3}{16}, \frac{3}{4}\right)$

Solution

Let $P\left(t_1^2, 2 t_1\right)$ and $Q\left(t_2^2, 2 t_2\right)$ are extremetre of chord
$$ \therefore \quad h=\frac{3 t_1^2+t_2^2}{4}, k=\frac{6 t_1+2 t_2}{4} $$ Slope of $P Q=2$ $$ \begin{array}{rlrl} \therefore & 2 & =\frac{2 t_1-2 t_2}{t_1^2-t_2^2} \\ t_1+t_2 & =1 \Rightarrow t_2=1-t_1 \end{array} $$ Put the value of $t_2$ in $h$ and $k$, we get $$ \begin{aligned} & 4 h=3 t_1^2+\left(1-t_1\right)^2, 4 k=6 t_1+2-2 t_1 \\ & 4 h=4 t_1^2-2 t_1+1,4 k=4 t_1+2 \end{aligned} $$ Eliminating $t_1$, we get $$ \begin{aligned} 4 h & =4\left(\frac{2 k-1}{2}\right)^2-2\left(\frac{2 k-1}{2}\right)+1 \\ 4 h & =4 k^2-6 k+3 \\ \Rightarrow\left(k-\frac{3}{4}\right)^2 & =4\left(h-\frac{3}{16}\right) \end{aligned} $$ $$ \therefore \operatorname{Vertex}\left(\frac{3}{16}, \frac{3}{4}\right) $$

Hence, the correct answer is option: 'D'.