Question

Find the locus of a point which divides the chord of slope 2 of the parabola $y^2=4ax$ in the ratio $1:2$ internally.

Answer 1

Parabola: $y^2=4ax...............\left(1\right)$

Let, end point of chord be $P(t_1^2,2t_1)$ and $Q(t_2^2,2t_2)$

Slope of line joining $P$ and $Q=\frac{2t_2-2t_1}{t_2^2-t_1^2}=2$ (given)

$=>\frac{1}{t_2+t_1}=1$

$=>t_1+t_2=1.................\left(2\right)$

Let there be a point $A(h,k)$ which divides $PQ$ in ratio $2:1$

$=>A(h,k)=(\frac{(1-{t_1)}^2+2t_1^2}{3},\frac{2(1-t_1)+4t_1}{3})$ (as $t_2=1-t_1$)

$=>A(h,k)=(\frac{1+3t_1^2-2t_2}{3},\frac{2+2t_1}{3})$

Eliminating $\left(t_1\right)=>h=\frac{1+3({\frac{3k-2}{2})}^2-2(\frac{3k-2}{2})}{3}$

$=>4h=9k^2-16k+8$

For locus $h\to x$ and $k\to y$

$=>9y^2-16y+8-4x=0.$