Question

Find the equation of the tangent and normal to the parabola $x^2-4x-8y+12=0$ at $\left(4,\frac{3}{2}\right)$.

Answer 1

$x^2-4x-8y+12=0$

Differentiating w.r.t $x$ we get

$2x-4-8\frac{dy}{dx}+0=0$

$\Rightarrow 8\frac{dy}{dx}=2x-4$

$\Rightarrow \frac{dy}{dx}=\frac{2x-4}{8}=\frac{x-2}{4}$

$\Rightarrow {\left[\frac{dy}{dx}\right]}_{(4,\frac{3}{2})}=\frac{4-2}{4}=\frac{2}{4}=\frac{1}{2}$

$\therefore$ Slope $m=\frac{1}{2}$

Equation of the tangent is $y-y_1=m(x-x_1)$

$y-\frac{3}{2}=\frac{1}{2}(x-4)$

$\Rightarrow \frac{2y-3}{2}=\frac{x-4}{2}$

$\Rightarrow 2y-3=x-4$

or $x-2y-1=0$ is the equation of the tangent.

Equation of the normal is $y-y_1=\frac{-1}{m}(x-x_1)$

$y-\frac{3}{2}=\frac{-1}{\frac{1}{2}}(x-4)$

$\Rightarrow \frac{2y-3}{2}=-2x+8$

$\Rightarrow 2y-3=-4x+16$

or $2y-3+4x-16=0$

or $4x+2y-19=0$ is the equation of the normal.