find the points at which the graph of 4x^2 - 4x + 12y^2 -6y = 10 has a vertical and horizontal tangent line.

Question

For the vertical only solve for y and for the horizontal only solve for x.

I got the deriviative to be:
(8x-4)/(-24y + 6).

For the horizontal, x = 1/2
For the vertical y = 1/4

I have to plug these into the original equation to get the points and solve for the rest of the equation but each time i do it, it doesn't turn out right.
Can someone check over my work and show me how to finish the problem?

Answer 1

I agree with your dy/dx and your solution for x = 1/2 and y = 1/4

so let's try x = 1/2
1 - 2 + 12y^2 - 6y = 10
12y^2 - 6y - 11 = 0
x = (6 ± √564)/24
= (3±√141)/12

so the two points with horizontal tangents are (1/2,(3+√141)/12) and
(1/2,(3-√141)/12)

Just repeat for the other value carefully substituting and solving using the quadratic equation