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

$$\begin{gathered} \mathrm{Given} \mathrm{parabola} {\mathrm{y}}^2=4\mathrm{x} \\ \Rightarrow \mathrm{a}=1 \\ \mathrm{Let} \mathrm{A}=({{\mathrm{t}}_1}^2,2{\mathrm{t}}_1),\mathrm{B}=({{\mathrm{t}}_2}^2,2{\mathrm{t}}_2) \mathrm{be} \mathrm{end} \mathrm{points} \mathrm{of} \mathrm{chord} \\ \mathrm{Given} \mathrm{that} \mathrm{m}=\frac{2\cancel{\mathrm{a}}\left(\cancel{{\mathrm{t}}_2-{\mathrm{t}}_1}\right)}{\cancel{\mathrm{a}}\left(\cancel{{\mathrm{t}}_2-{\mathrm{t}}_1}\right)({\mathrm{t}}_2+{\mathrm{t}}_1)}=2 \\ \Rightarrow {\mathrm{t}}_1+{\mathrm{t}}_2=1 \\ Let P(x,y) be locus of point which divides AB in the ratio 1:3 then \\ \mathrm{P}(\mathrm{x}, \mathrm{y})=\left(\frac{{\mathrm{t}}_2^2+3{\mathrm{t}}_1^2}{4},\frac{2{\mathrm{t}}_2+6{\mathrm{t}}_1}{4}\right) \end{gathered}$$

$$\begin{gathered} 4\mathrm{x}={\left(1-{\mathrm{t}}_1\right)}^2+3{{\mathrm{t}}_1}^2, 4\mathrm{y}=2(1-{\mathrm{t}}_1)+6{\mathrm{t}}_1 \\ \Rightarrow 4\mathrm{x}=1-2{\mathrm{t}}_1+4{{\mathrm{t}}_1}^2 , 4\mathrm{y}=2+4{\mathrm{t}}_1 \\ \Rightarrow 4\mathrm{x}=1-2\left(\frac{4\mathrm{y}-2}{4}\right)+4{\left(\frac{4\mathrm{y}-2}{4}\right)}^2 \\ \Rightarrow 4\mathrm{x}=1-2\mathrm{y}+1+4{\mathrm{y}}^2+1-4\mathrm{y} \\ 4{\mathrm{y}}^2-6\mathrm{y}-4\mathrm{x}+3=0 \\ for this parabola \\ \therefore \mathrm{Vertex} =\left(\frac{3}{16},\frac{3}{4}\right) \end{gathered}$$