Find the equation of the tangent line to the curve 2x^2y-3y^2=-11 at the point (2,-1).

Answer 1

2x^2y-3y^2=-11
4xy + 2x^2 dy/dx - 6y dy/dx = 0
dy/dx(2x^2 - 6y) = -4xy
dy/dx = -4xy/(2x^2 - 6y)
at (2,-1)
dy/dx = -4(2)(-1)/(8 -6(-1))
= 8/14 = 4/7

equation of tangent:
y+1 = (4/7)(x - 2)

massage it any way you need to