The normal at (2, (3)/(2)) to the ellipse, (x^2)/(16)+(y^2)/(3)=1...

Ellipse is $\frac{x^2}{16}+\frac{y^2}{3}=1$ Now, equation of normal at $(2,3 / 2)$ is $$\begin{aligned}& \frac{16 x}{2}-\frac{3 y}{3 / 2}=16-3 \\ & \Rightarrow \quad 8 x-2 y=13 \\ & \Rightarrow \quad y=4 x-\frac{13}{2} \end{aligned}$$Let $y=4 x-\frac{13}{2}$ touches a parabola $y^2=4 a x$. We know, a straight line $y=m x+c$ touches a parabola $y^2=4 a x$ if $a-m c=0$ \therefore \quad a-(4)$\left(-\frac{13}{2}\right)=$0 \Rightarrow a=-26 Hence, required equation$ of parabola isy^2=4(-26) x=-104 x