Parametrization of the intersection of two given surfaces

Question

Find a parametrization of the intersection between the two curves $z=x^2-y^2$ and $z=x^2+xy-1$.

I figure I should set them equal to each other but I'm not sure where to go from there: $$x^2-y^2 = x^2+xy-1$$

Answer 1

$$ x^2 - y^2 = x^2 + xy - 1 \\ 0 = y^2 + xy - 1 $$

Apply the quadratic formula over $y$.

$$ y = \frac{x \pm \sqrt{x^2 - 4(1)(-1)}}{2(1)} = \frac{x \pm \sqrt{x^2 + 4}}{2} $$

Alternatively, we can solve for $x$.

$$ 0 = y^2 + xy - 1 \\ x = \frac{1 - y^2}{y} $$

From there, you can continue.

Answer 2

Summary of comments:

1. Set them equal to each other, $x^2-y^2 = x^2+xy-1$
2. Simplify the resulting equation: $0=y^2+xy-1$
3. The equation is easier to solve for $x$: $x=(1-y^2)/y$
4. Make $y=t$ as the parameter, obtaining $x=(1-t^2)/t$ and $y=t$, and also $z=x^2-y^2 = (1-t^2)^2/t^2 - t^2$.

Note that the intersection consists of two curves, corresponding to the $t<0$ and $t>0$.

Answer 3

Eliminate $ z - x^2 $ to get $ x = y -1/y $.

Let $ y = t, x = 1/t -t $, simplify to get $ z = 1/t^2 -2.$