Question

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

Answer 1

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

Explanation:

Equation of normal at `x=x_0` on the curve `y=f(x)` is perpendicular to the tangent at `x=x_0` at `f(x)`.

As slope of tangent at `x=x_0` on the curve `f(x)`, is given by `f'(x_0)`, slope of normal is `-1/(f(x_0))` and equation of normal is

`y=-1/(f(x_0))(x-x_0)+f(x_0)`

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

`2x-2y(dy)/(dx)=0` of `f'(x)=(dy)/(dx)=x/y`

Hence slope of tangent at `(5,3)` is `5/3` and slope of normal is `-3/5`.

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