Question

If $PQ$ is double ordinate of the parabola $y^2=4ax$ then locus of its point of trisection is

• A. $9y^2=4ax$
• B. $9y^2=16ax$
• C. $3y^2=8ax$
• D. $9y^2=8ax$

Answer 1

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

Parametric point of parallel $y^2=4ax$ is $(a{t}^2,2at)$

A double ordinate is a line parallel to axis of parabola.

Hence, if $A$ and $B$ are ends of double ordinate then :

$A(a{t}^2,2at)$ and $B(a{t}^2,-2at)$

Let $C(h,k)$ be point of trisection. i.e. $C$ divides $AB$ in ratio $1:2$.

By section formula we have:

$(h,k)=(\frac{1\times a{t}^2+2\left(a{t}^2\right)}{1+2},\frac{1(-2at)+2\left(2at\right)}{1+2})$

$(h,k)=(\frac{3a{t}^2}{3},\frac{4at-2at}{3})$

$(h,k)=(a{t}^2,\frac{2at}{3})$

Now,

$h=a{t}^2$

$k=\frac{2at}{3}$

So, $2at=3k$

$\therefore$ The value of $(a{t}^2,2at)$ becomes $(h,3k)$ which is lies on parabola.

$y^2=4ax$

$(3k{)}^2=4ah$

$\Rightarrow$ $9k^2=4ah$

Now, substituting $h=x$ and $k=y$

$\Rightarrow$ $9y^2=4ax$