Use implicit differentiation to find an equation of the tangent line to the curve at the given point.

x^2+2xy-y^2+x=6 , (2,4) (hyperbola)

2 answers

f'(x) =
2x +2y + 2xdy/dx - 2ydy/dx + 1 = 0

2xdy/dx - 2ydy/dx = -2x - 2y -1

dy/dx = (-2x - 2y - 1)/ (2x - 2y)

the tangent passes through (2,4)

m = dy/dx = (-2(2) - 2(4) -1 )
------------------
(2(2) - 2(4))

m = 13/4

the equation of the tangent line through (2,4);
y = mx + c
4 = 13/4(2) + c
c = -5/2

y = 13/4x - 5/2
thanks for the help.