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

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Answer 1 · 2017-02-27T09:59:15

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

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]}