I have two problems dealing with this. I am getting myself all screwed up and could use help.
the first is at what points is the slope of the tangent line equal to 1 on [x^2/9] + [y^2/16] = 1
So I got the derivative dy/dx = -[16x/9y]
so now I solve for -[16x/9y] = 1 ; correct?
The other question is find the points at which the tangent is horizontal for
x^3 + 4x^2 +xy^2 - 4y^2 = 0
I got dy/dx = [3x^2 + 8x +y^2]/6
so same thing, solve for dy/dx to be zero?
Can you show me how you would solve to get that, I'm not getting it. And I tried using wolfram but it's losing me. It's saying multiple both sides by the denominator to clear the fraction. Which I understand in other problems fine but it's confusing me with this.
should have mentioned there was a diagram showing that the x value wouldn't be exceeding 4 for the the right value for x is x = 2 - 2sqrt5
so plugging that in I get
y^2 = -3[2-2sqrt5] - 8[2 - 2sqrt5]
distribute those out for y = sqrt[-22 + 22sqrt5] but thats not the right answer. what am I missing?