Question

The locus of a point divides a chord of slope $2$ of the parabola $y^2=4x$ internally in the ratio $1:2$ is

• A. $(y+\frac{8}{9}{)}^2=\frac{4}{9}(x-\frac{2}{9})$
• B. $(y-\frac{8}{9}{)}^2=\frac{4}{9}(x-\frac{2}{9})$
• C. $(y-\frac{8}{9}{)}^2=\frac{4}{9}(x+\frac{2}{9})$
• D. $(y+\frac{8}{9}{)}^2=\frac{4}{9}(x+\frac{2}{9})$

Answer 1

The correct option is B $(y-\frac{8}{9}{)}^2=\frac{4}{9}(x-\frac{2}{9})$

Locus of point a chord of slope $2$ of parabola $y^2=4x$

Internally in ratio $1:2$ is:

Let two parabola be $(t_1^2,2t_1)$ and $(t_2^2,2t_2)$

Slope of line $=\frac{2t_2-2t_1}{t_2^2-t_1^2}=2$

$=t_1+t_2=1$

Hence two points are $(t_1^2,2t_1)$ and $\left(\right(1-t_1^2),2(1-t_1\left)\right)$

Let $(h,k)$ be point which divides line in $1:2$

$h=\frac{(1-t_1{)}^2+2t_1^2}{1+2}=\frac{3t_1^2-2t_1+1}{3}k=\frac{2(1+t_1)+4t_1}{1+2}=\frac{2t_1+2}{3}$

Solve $h$ and $k$ to establish $t_1$

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

Replace $h\to xk\to y$

${(y-\frac{8}{9})}^2=\frac{4}{9}(x-\frac{2}{9})$