To find the slope of the tangent line to the curve at a given point, we first need to differentiate the equation implicitly. The given curve is:
3y² + 2x² – 6 – 2xy = 0
We differentiate both sides with respect to x:
Using implicit differentiation:
- Differentiate 3y²: 6y(dy/dx)
- Differentiate 2x²: 4x
- Differentiate -6: 0
- Differentiate -2xy using the product rule: -2(y + x(dy/dx))
Putting it all together, we get:
6y(dy/dx) + 4x – 2(y + x(dy/dx)) = 0
Now, simplify the equation:
6y(dy/dx) – 2x(dy/dx) + 4x – 2y = 0
Factor out dy/dx:
(6y – 2x)(dy/dx) + 4x – 2y = 0
Solving for (dy/dx):
(dy/dx) = (2y – 4x) / (6y – 2x)
Now we substitute the point (3, 2) into this derivative:
dy/dx = (2(2) – 4(3)) / (6(2) – 2(3))
Calculating this gives:
dy/dx = (4 – 12) / (12 – 6) = -8 / 6 = -4/3
Thus, the slope of the tangent line to the curve at the point (3, 2) is:
-4/3