A curve has the equation $ x=2t^{2} $ and $ y=3t $ and a circle has the equation $ x^{2} + y^{2}-6x-1 =0 $
What are the coordinates of the intersections between the objects?
I tried subbing the x and y equations into the circle formula. This produced $ 4t^{4} + 9t^{2} -12t^{2} -1 =0 $ However, I could not solve for t and hence could not sub to get the coordinates of the points.
What is the best way to tackle this question and is there a better way to solve these type of questions?
You should have gotten $$4t^4 + 9t^2 - 12t^2 - 1 = 4t^4 - 3t^2 -1=0$$
Now, put $u = t^2$ to get $$4u^2 - 3u - 1 = (4u +1)(u-1)=0 $$ so $u = -\frac 14$ or $u = 1$ and then, since $t = \pm \sqrt u$, we need to throw out $t=\pm \sqrt{-\frac 14}$ as a solution in the reals, so we are left with $t = \pm 1$.
Substitute these values of $t$ into the parametric equations to find the $x, y$ coordinates of the points of intersection: $(x, y) = (2t^2, 3t)$.
This is a fine procedure to use, and works well once you've used it a time or two.