Question:
Find the equation of the tangent and the normal to the following curves at the indicated points:
$x^{2}=4 y$ at $(2,1)$
Solution:
finding the slope of the tangent by differentiating the curve
$2 \mathrm{x}=4 \frac{\mathrm{dy}}{\mathrm{dx}}$
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{x}}{2}$
$\mathrm{m}($ tangent $)$ at $(2,1)=1$
normal is perpendicular to tangent so, $m_{1} m_{2}=-1$
$\mathrm{m}($ normal) at $(2,1)=-1$
equation of tangent is given by $y-y_{1}=m(\operatorname{tangent})\left(x-x_{1}\right)$
$y-1=1(x-2)$
equation of normal is given by $y-y_{1}=m($ normal $)\left(x-x_{1}\right)$
$y-1=-1(x-2)$
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