Question

If $Q{Q}^{\prime }$ is a double ordinate of a parabola $y^2=9x$,then the locus of its point of trisection is:

• A. $y^2=x$
• B. $y^2=3x$
• C. $y^2=6x$
• D. None of these

Answer 1

The correct option is A $y^2=x$

Let $P(h,k)$ be the point which trisect the double ordinate $Q{Q}^{\prime }$

Coordinates of $Q$ are $(a{t}^2,2at)$ and $Q^{\prime }$ are $(a{t}^2,-2at)$

Coordinates of point which divides $Q{Q}^{\prime }$ in $1:2$ are

$P(\frac{a{t}^2\left(1\right)+a{t}^2\left(2\right)}{2+1},\frac{-2at\left(1\right)+2at\left(2\right)}{2+1})\Rightarrow P(a{t}^2,\frac{2at}{3})h=a{t}^2......\left(i\right)k=\frac{2at}{3}\Rightarrow t=\frac{3k}{2a}$

Substituting $t$ in $\left(i\right)$

$h=a{\left(\frac{3k}{2a}\right)}^{2}h=a\frac{9k^2}{4a^2}\Rightarrow 9k^2=4ah$

Replacing $h$ by $x$ and $k$ by $y$

$\Rightarrow$ $9y^2=4ax$ is the locus. --- $\left(ii\right)$

The equation of parabola $y^2=9x$ comparing it with $y^2=4ax$

$\therefore$ We get, $a=\frac{9}{4}$

Substituting value of $a$ in $\left(ii\right)$ we get,

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

$\therefore$ $y^2=x$

$\therefore$ Te locus of its point of trisection is $y^2=x$