# Thread: How would you find the horizontal and vertical tangents of (x^2+y^2)^2=x^2-y^2?

1. ## How would you find the horizontal and vertical tangents of (x^2+y^2)^2=x^2-y^2?

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?