The question reads as follows:
Find a parametrization of the curve resulting from the intersection of the surfaces:
$z = x^2 – y^2$ and $z= x^2 +xy – 1$
My attempt:
(Use y = t as a parameter, so solve for y)
$z = x^2-y^2$ and $y^2 = \frac{(z-x^2+1)^2}{x^2}$
so we have $x^2y^2 = (-y^2+1)^2$
$xy = -y^2+1$ then $x = -y + \frac{1}{y}$
then we also have $z = (\frac{-y+1}{y})^2 – y^2$
so $z = \frac{y^2 -2y + 1}{y^2} – y^2$
$z = 1 – \frac{2}{y} + \frac{1}{y^2} – y^2$
now let $y = t$ be the parameter so we have:
$\langle – t + \frac{1}{t},\ t, \ 1 – \frac{2}{t} + \frac{1}{t^2} – t^2 \rangle$
However, I have no idea if this is a [Deleted: the] correct parametric curve. Any thoughts?
Edit: after looking over my working I realized I made an algebra mistake. I have checked the following parametric equation and it seems correct:
$\langle – t + \frac{1}{t},\ t, \ – 2 + \frac{1}{t^2} \rangle$
Best Answer
The word "the" is not appropriate, here. Any given curve can have many different sets of parametric equations. So, the question is whether this is a correct parametric equation.
To check that your answer is correct, just substitute back in the equations of the two surfaces. The point $\big(x(t), y(t), z(t)\big)$ given by your parametric equations should satisfy both surface equations for all values of $t$.
There's another slightly subtle point, though ...
If you do what I said above, then you have shown that your equations represent some subset of the intersection, but maybe not all of it. It quite often happens that the intersection of two surfaces consists of two or more disconnected pieces. For example, think of the two loops formed when a small cylinder intersects a larger one at right angles. Any given set of parametric equations can represent one of the pieces, but not more than one. So, to be thorough, you really ought to show that every point on the intersection will be given by some value of $t$ in your equations. But this is maybe more than you were expected to do in this question.