Question:
Find the equation of the tangent and the normal to the following curves at the indicated points:
$y^{2}=\frac{x^{3}}{4-x}$ at $(2,-2)$
Solution:
finding the slope of the tangent by differentiating the curve
$2 y \frac{d y}{d x}=\frac{(4-x) 3 x^{2}+x^{4}}{(4-x)^{2}}$
$\frac{d y}{d x}=\frac{(4-x) 3 x^{2}+x^{4}}{2 y(4-x)^{2}}$
$m$ (tangent) at $(2,-2)=-2$
$\mathrm{m}$ (normal) at $(2,-2)=\frac{1}{2}$
equation of tangent is given by $y-y_{1}=m(t a n g e n t)\left(x-x_{1}\right)$
$y+2=\frac{1}{2}(x-2)$
$2 y+4=x-2$
$2 y-x+6=0$
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