Find the equation of the tangent and normal to the parabola `y^2=4a x` at the point `(a t^2,\ 2a t)` .

Answer

`y=-t x+2 a t+a t^{3}`

Explanation

Given, equation of parabola `y^{2}=4 ax`

Differentiating w.r.t. `x`,
$$ \begin{aligned} &2 y \frac{\mathrm{dy}}{\mathrm{dx}}=4 \mathrm{a} \\ &\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2 \mathrm{a}}{y} \end{aligned} $$
...