Question

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

• A. ($\frac{1}{9},\frac{2}{9}$)
• B. ($\frac{8}{9},\frac{1}{9}$)
• C. ($\frac{2}{9},\frac{8}{9}$)
• D. ($\frac{1}{9},\frac{1}{9}$)

Answer 1

The correct option is C ($\frac{2}{9},\frac{8}{9}$)

Let the two points on the given parabola be $(t_1^2,2t_1)$ & $(t_2^2,2t_2)$

Slope of the line joining these points is

$2=\frac{t_2^2-2t_1}{t_1^2-t_1^2}=\frac{2}{t_1+t_2}$

$\Rightarrow t_1+t_2=1$

Hence two points become $(t_1^2,2t_1)$

& $((1-t_1^2),2(1-t_1))$

Let $(h,k)$ be the point which divides these points in the ratio $1:2$

$h={(1-t_1)}^2,2\frac{t_1}{3}$

$=1-2t_1+3\frac{t_1^2}{3}\to 1$

$k=[2(1-t_1)+4t_1]/3=2+2\frac{t_1}{3}\to 2$

Eliminating $t_1$ from $1$ and $2$

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

Hence the locus of $(h,k)$ is

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

This is a parabola with vertex $(\frac{2}{9},\frac{8}{9})$