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Q.
The equations of the tangent and normal to the parabola $y^2=4 a x$ at the point $\left(a t^2, 2 a t\right)$ are respectively
Application of Derivatives
Solution:
The equation of the given parabola is $y^2=4 a x ......$(i)
On differentiating w.r.t. $x$, we get
$2 y \frac{d y}{d x}=4 a \Rightarrow \frac{d y}{d x}=\frac{2 a}{y}$
$\therefore$ Slope of the tangent at $\left(a t^2, 2 a t\right)$ is
$\left(\frac{d y}{d x}\right)_{\left(a t^2, 2 a t\right)}=\frac{2 a}{2 a t}=\frac{1}{t}$
Hence, the equation of the tangent at $\left(a t^2, 2 a t\right)$ is
$y-2 a t =\frac{1}{t}\left(x-a t^2\right) $
$\Rightarrow y t-2 a t^2=x-a t^2 \Rightarrow x-t y+a t^2=0$
Also, slope of the normal at
$\left(a t^2, 2 a t\right)=\frac{-1}{\text { Slope of tangent }} a t\left(a t^2, 2 a t\right)=-t$
$\therefore$ The equation of the normal at $\left(a t^2, 2 a t\right)$ is
$y-2 a t=-t\left(x-a t^2\right) \Rightarrow t x+y-2 a t-a t^3=0$