Find the equation of the tangent and the normal to the following curves at the indicated points:
$x=a t^{2}, y=2$ at at $t=1$
finding slope of the tangent by differentiating $x$ and $y$ with respect to $t$
$\frac{\mathrm{dx}}{\mathrm{dt}}=2 \mathrm{at}$
$\frac{\mathrm{dy}}{\mathrm{dt}}=2 \mathrm{a}$
Now dividing $\frac{\mathrm{dy}}{\mathrm{dt}}$ and $\frac{\mathrm{dx}}{\mathrm{dt}}$ to obtain the slope of tangent
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\mathrm{t}}$
$\mathrm{m}$ (tangent) at $\mathrm{t}=1$ is 1
normal is perpendicular to tangent so, $m_{1} m_{2}=-1$
$\mathrm{m}($ normal $)$ at $\mathrm{t}=1$ is $-1$
equation of tangent is given by $y-y_{1}=m($ tangent $)\left(x-x_{1}\right)$
$y-2 a=1(x-a)$
equation of normal is given by $y-y_{1}=m($ normal $)\left(x-x_{1}\right)$
$y-2 a=-1(x-a)$
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