Question

The locus of points of trisection of double ordinate of $y^2=4ax$ is

• A. $y^2=ax$
• B. $9y^2=4ax$
• C. $9y^2=ax$
• D. $y^2=9ax$

Answer 1

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

Let Co-ordinates of double ordinate to parabola $y^2=4ax$ are $(a{t}^2,2at)$ and $(a{t}^2,-2at)$

So, points of trisection are $(\frac{a{t}^2\left(1\right)+a{t}^2\left(2\right)}{1+2},\frac{2at\left(1\right)+(-2at)\left(2\right)}{1+2})$
and $(\frac{a{t}^2\left(2\right)+a{t}^2\left(1\right)}{2+1},\frac{2at\left(2\right)+(-2at)\left(1\right)}{2+1})$
$\equiv (a{t}^2,\frac{-2at}{3})\text{and}(a{t}^2,\frac{2at}{3})\equiv (h,k)\text{Let}$

$\Rightarrow t=\pm \frac{3k}{2a}$ and $h=a{t}^2$
$\Rightarrow h=a\left(\frac{9k^2}{4a^2}\right)$
$\Rightarrow h=\frac{9k^2}{4a}$
$\Rightarrow 9y^2=4ax$