Find the slope of the line tangent to the curve at each point on the curve where x=6

Question

Curve equation 2x^2+3y^2−4xy=36.

Answer 1

2x^2+3y^2−4xy=36

Solve for y when x=6

2(6^2) + 3y^2 - 4(6)y = 36

72 + 3y^2 - 24y =36

3y^2 - 24y + 36 = 0

y^2 - 8y + 12 = 0

(y-6)(y-2) = 0

The two points where x=6 are (6,6) and (6,2)

We need the derivative dy/dx. This will require implicit differentiation.

4x + 6y(dy/dx) - [ 4x(dy/dx) + 4y ] = 0

Group together the terms with dy/dx in them

4x + 6y(dy/dx) - 4x(dy/dx) - 4y = 0

Isolate the terms with dy/dx in them

6y(dy/dx) - 4x(dy/dx) = 4y - 4x

Factor out (dy/dx)

(dy/dx) [6y - 4x] = 4y - 4x

dy/dx = (4y - 4x) / (6y - 4x)

dy/dx at the point (6,6) = 0

dy/dx at the point (6,2) = (8 - 24) / (12 - 24) = 4/3