Find the equations of the tangent and normal to the curve $y^2 = \dfrac{x^3}{4-x}$ at the point $\left(2,-2\right)$
$y^2 = \dfrac{x^3}{4-x}$
Differentiating w.r.t x gives
$2y \dfrac{dy}{dx} = \dfrac{\left(4-x\right)\times 3x^2 -x^3 \times \left(-1\right)}{\left(4-x\right)^2}$
i.e. $\dfrac{dy}{dx} = \dfrac{1}{2y} \left[\dfrac{12x^2-2x^3}{\left(4-x\right)^2}\right]$
i.e. $\dfrac{dy}{dx} = \dfrac{1}{y} \left[\dfrac{6x^2 -x^3}{\left(4-x\right)^2}\right]$
$\therefore$ $\;$ $\dfrac{dy}{dx} \bigg|_{\left(2,-2\right)} = \dfrac{-1}{2}\left[\dfrac{\left(6\times 2^2\right)-2^3}{\left(4-2\right)^2}\right]$
i.e. $\dfrac{dy}{dx} \bigg|_{\left(2,-2\right)} = \dfrac{-1}{2} \times \dfrac{16}{4}=-2$
$\therefore$ $\;$ Slope of tangent at $\left(2,-2\right) = -2$
$\therefore$ $\;$ Equation of tangent at $\left(2,-2\right)$ is $\;\;$ $y+2 = -2 \left(x-2\right)$
i.e. $2x+y-2=0$
Slope of normal at $\left(2,-2\right)$ $= \dfrac{-1}{\text{slope of tangent}} = \dfrac{1}{2}$
$\therefore$ $\;$ Equation of normal at $\left(2,-2\right)$ is $\;\;$ $y+2 = \dfrac{1}{2} \left(x-2\right)$
i.e. $x-2y-6=0$