# How do you determine the intersection between parametric curves (t^2+2t,-3t^2+5t) and (-2t^2+4t,t^2+2t)?

## As mentioned in the title, I have the following parametric curves: Red: $\left({t}^{2} + 2 t , - 3 {t}^{2} + 5 t\right)$ Blue: $\left(- 2 {t}^{2} + 4 t , {t}^{2} + 2 t\right)$ I have tried to write the curves in function form using the method that PatrickJMT demonstrates, but I encountered two $\pm$ symbols in the same equation, and I am unsure how to proceed. How do I find the exact intersections? Thanks!

Oct 8, 2016

See below.

#### Explanation:

Defining

${f}_{1} \left(x \left(t\right) , y \left(t\right)\right) = \left\{{t}^{2} + 2 t , - 3 {t}^{2} + 5 t\right\}$ and
${f}_{2} \left(x \left(t\right) , y \left(t\right)\right) = \left\{- 2 {t}^{2} + 4 t , {t}^{2} + 2 t\right\}$

We can verify that the set of coincidence points only has one element: The point $\left(0 , 0\right)$ at the instant $t = 0$.

The functions

${f}_{1} \left(x , y\right)$ and ${f}_{2} \left(x , y\right)$ have crosses in two points:

$\left\{0 , 0\right\}$ and $\left\{1.8636 , 2.02359\right\}$

Those points are the solutions of the nonparametric curves

${f}_{1} \left(x , y\right) = \sqrt{1 + x} - \frac{1}{6} \left(5 - \sqrt{25 - 12 y}\right) - 1 = 0$ and
${f}_{2} \left(x , y\right) = \frac{1}{2} \left(2 - \sqrt{2} \sqrt{2 - x}\right) - \left(\sqrt{1 + y} - 1\right) = 0$