Find the equations of the tangent and normal to the parabola $y^{2} = 4 a x$ at the point ( $a t^{2}, 2 a t$ ).

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Solution

Given, equation of parabola is $y^{2} = 4 a x$
Differentiating w.r.t. $x$ , we get
$2 y \frac{d y}{d x} = 4 a$
$\Rightarrow \frac{d y}{d x} = \frac{2 a}{y}$
Slope of the tangent at $(a t^{2}, 2 a t)$ is
${(\begin{matrix} \frac{d y}{d x} \end{matrix})}_{(a t^{2}, 2 a t)} = \frac{2 a}{2 a t} = \frac{1}{t}$

Equation of the tangent at $(a t^{2}, 2 a t)$ is given by,
$y - 2 a t = \frac{1}{t} (x - a t^{2})$
$\Rightarrow t y - 2 a t^{2} = x - a t^{2}$
$\Rightarrow t y = x + a t^{2}$

Slope of normal at $(a t^{2}, 2 a t)$ is given by,
$\frac{- 1}{Slope of the tangent at (a t^{2}, 2 a t)} = - t$

Equation of the normal at $(a t^{2}, 2 a t)$ is given by
$y - 2 a t = - t (x - a t^{2})$
$\Rightarrow y - 2 a t = - t x + a t^{3}$
$\Rightarrow y = - t x + 2 a t + a t^{3}$

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