# Find the equation

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 $y-y_{1}=m(t a n g e n t)\left(x-x_{1}\right)$

$y+2=\frac{1}{2}(x-2)$

$2 y+4=x-2$

$2 y-x+6=0$