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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?

Tomato721
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    @amWhy has given the main answer, but to the second question: what you did -- reducing to a one-dimensional problem -- is a generally good way to go. You just have to get the algebra right. :) – John Hughes Jan 20 '15 at 14:56
  • Yeah, I did the algebra like I was supposed to but I mis-typed it here (first time using latex). I didn't know how to solve the resulting quadratic though. – Tomato721 Jan 20 '15 at 15:05

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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.

amWhy
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