Question

$PQ$ is a double ordinate of the parabola $y^2=4ax$. The locus of the points of trisection of $PQ$ is ?

• A. $9y^2=4ax$
• B. $9x^2=4ay$
• C. $9y^2+4ax=0$
• D. $9x^2+4ay=0$

Answer 1

The correct option is A $9y^2=4ax$

Here, from the equation $y^2=4ax$ we can write,

$x=a{t}^2,y=2at$

If the extremities of the double ordinate are $(a{t}^2,2at)$ and $(a{t}^2,-2at)$, then its points of trisection have ordinates;

$y_1=\frac{-2at+4at}{3}=\frac{2at}{3}{y}_2=\frac{-4at+2at}{3}=\frac{-2at}{3}$

Taking any point,

$y_1=\frac{2at}{3}\Rightarrow y_1^2=\frac{4a^2{t}^2}{9}\Rightarrow y_1^2=\frac{4a\left(a{t}^2\right)}{9}$

Comparing $y_1$ to $y$ and $a{t}^2$ to $x$,

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

Hence, the required locus of the points of trisection of $PQ$ is $9y^2=4ax.$