Question

PQ is a double ordinate of a parabola. Find the locus of its points of trisection.

Answer 1

Let the equation of parabola be $y^2=4ax.$ Extremities of double ordinate in parametric form are $(a{t}^2,2at);(a{t}^2,-2at)$

Let $(x_1,y_1)\&(x_2,y_2)$ be points of trisections .

$\therefore x_1=x_1=a{t}^2$

$y_1=-2at+\frac{4at}{3}=\frac{-2at}{3}$

$y_2=-2at+\frac{8at}{3}=\frac{2at}{3}$

$\therefore y_1^2=\frac{4}{9}{a}^2{t}^2\&y_2^2=\frac{4}{9}{a}^2{t}^2$

$\Rightarrow$ Generalized point $\Rightarrow y^2=\frac{4ax}{9}$

Hence, locus of point of trisection is given by $y^2=\frac{4}{9}ax.$