Conic Sections — Maths STD 11 Science — Question

$\Rightarrow y^2=\frac{16}{3} x$

Comparing this equation with $y^2=4 a x$, we get

$\begin{aligned} & \Rightarrow 4 a=\frac{16}{3} \\ & \Rightarrow a=\frac{4}{3}\end{aligned}$

If t is the parameter of the point P on the parabola, then

$\begin{aligned} & P(t)=\left(a t^2, 2 a t\right) \\ & \text { i.e., } x=a t^2 \text { and } y=2 a t\end{aligned}$

………(i)

(i) Given point is (3, -4)

Substituting $x=3, y=-4$ and $a=\frac{4}{3}$ in (i), we get

$\begin{aligned} & 3=\frac{4}{3} t^2 \text { and }-4=2\left(\frac{4}{3}\right) t \\ & t^2=\frac{9}{4} \text { and } t=\frac{-3}{2} \\ & t= \pm \frac{3}{2} \text { and } t=\frac{-3}{2} \\ & t=-\frac{3}{2}\end{aligned}$

$\therefore$ The parameter of the given point is $\frac{-3}{2}$