What is the equation of the tangent line to the curve 2x^2y+3y^2=4x+6y at the point (0,2)?

First we calculate slope at point (0,2) by finding dy/dx (using implicit differentiation)

2x²y + 3y² = 4x + 6y

4xy + 2x² dy/dx + 6y dy/dx = 4 + 6 dy/dx
dy/dx (2x² + 6y - 6) = 4 - 4xy
dy/dx = 4(1 - xy) / (2(x² + 3y - 3))
dy/dx = 2 (1 - xy) / (x² + 3y - 3)

At point (0,2), dy/dx = 2(1-0)/(0+6-3) = 2/3

Tangent line has slope 2/3 and passes through point (0, 2)
y - 2 = 2/3 (x - 0)
y = 2/3 x + 2

Mαthmφm

First we calculate slope at point (0,2) by finding dy/dx (using implicit differentiation)

2x²y + 3y² = 4x + 6y

4xy + 2x² dy/dx + 6y dy/dx = 4 + 6 dy/dx
dy/dx (2x² + 6y - 6) = 4 - 4xy
dy/dx = 4(1 - xy) / (2(x² + 3y - 3))
dy/dx = 2 (1 - xy) / (x² + 3y - 3)

At point (0,2), dy/dx = 2(1-0)/(0+6-3) = 2/3

Tangent line has slope 2/3 and passes through point (0, 2)
y - 2 = 2/3 (x - 0)
y = 2/3 x + 2

Mαthmφm