# How do you find the equations for the normal line to x^2-y^2=16 through (5,3)?

Feb 27, 2017

Equation of normal is $3 x + 5 y - 30 = 0$

#### Explanation:

Equation of normal at $x = {x}_{0}$ on the curve $y = f \left(x\right)$ is perpendicular to the tangent at $x = {x}_{0}$ at $f \left(x\right)$.

As slope of tangent at $x = {x}_{0}$ on the curve $f \left(x\right)$, is given by $f ' \left({x}_{0}\right)$, slope of normal is $- \frac{1}{f \left({x}_{0}\right)}$ and equation of normal is

$y = - \frac{1}{f \left({x}_{0}\right)} \left(x - {x}_{0}\right) + f \left({x}_{0}\right)$

As ${x}^{2} - {y}^{2} = 16$ taking derivative w.r.t. $x$, we get

$2 x - 2 y \frac{\mathrm{dy}}{\mathrm{dx}} = 0$ of $f ' \left(x\right) = \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x}{y}$

Hence slope of tangent at $\left(5 , 3\right)$ is $\frac{5}{3}$ and slope of normal is $- \frac{3}{5}$.

Hence equation of normal is $y - 3 = - \frac{3}{5} \left(x - 5\right)$ or $3 x + 5 y - 30 = 0$
graph{(3x+5y-30)(x^2-y^2-16)=0 [-7.125, 12.875, -2.68, 7.32]}