1
$\begingroup$

I've been working on converting parametric equations into Cartesian form, but can't figure this out.

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

How do I covert that to Cartesian? Any help would be most appreciated.

$\endgroup$
1
$\begingroup$

Since $(t^2 - 1)x = t^2 + 1$ and $(t^2 - 1)y = 2t$, $$(t^2 - 1)^2 x^2 - (t^2 - 1)^2 y^2 = (t^2 + 1)^2 - 4t^2 = (t^4 + 2t^2 + 1) - 4t^2 = t^4 - 2t^2 + 1 = (t^2 - 1)^2.$$

Dividing through by $(t^2 - 1)^2$, we obtain $$x^2 - y^2 = 1$$

$\endgroup$
0
$\begingroup$

In general, you need to eliminate $t$ with the 2 equations.

For this one, we get a shortcut.

$$x+y={t+1 \over t-1}$$ $$x-y={t-1 \over t+1}$$

Then, obviously,

$$(x+y)(x-y)=1$$ $$(x^2-y^2)=1$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.