Carrying on: 2x^2+2y^2=1 2(x^2+y^2)=1 (x^2+y^2)^2=1/4 Sub. in original x^2-y^2=1/4 x^2=1/4+y^2
Sub this in my 3rd equation (1/4+2y^2)^2=1/4 Hence get x and y (I get y^2=1/8 and x^2=3/8)
The way I understand it is that to find the horizontal tangents you equate the numerator to zero, and for the vertical tangents, I equate the denominator to 0. But, when I equate the numerator to 0, I get this: x = 0, and 2x^2 + 2y^2 -1 = 0, how do I go about solving this?
Also, the funny thing here is that when I graph (x^2+y^2)^2=x^2-y^2 out (it's a lemniscate graph), the tangent at x = 0 is not horizontal. Why is that?
Thanks for your help!
Carrying on: 2x^2+2y^2=1 2(x^2+y^2)=1 (x^2+y^2)^2=1/4 Sub. in original x^2-y^2=1/4 x^2=1/4+y^2
Sub this in my 3rd equation (1/4+2y^2)^2=1/4 Hence get x and y (I get y^2=1/8 and x^2=3/8)