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I stumbled on this one a few days ago and I'm probably missing something obvious...

I basically need to solve those parametric equations for the other coordinate $(x,y)$ other than the point $(t^2,t^3)$:

$$2y = 3tx - t^3$$ $$x = t^2$$ $$y = t^3$$

To have more context, the question started with $x = t^2$, $y = t^3$, asking to find the equation of the tangent at the point $(t^2,t^3)$, then to find the other point where this tangent meets the parametric curve.

It just seems that whenever I try something, I get back to $(t^2,t^3)$ instead of the other point. I've been trying with:

$2(t^3) = 3tx - t^3$, then
$2y = 3t(t^2) - t^3$, then
$2\sqrt{x^3} = 3tx - t^3$ (but this one is intimidating for something which should be much simpler than that)

Is there perhaps a way to exclude $(t^2,t^3)$ to get the other solution? The other point should be $\left(\dfrac{t^2}{4},-\dfrac{t^3}{8}\right)$ by the way.

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At first sight, your first equation $2y = 3tx - t^3$ does not bring out new information to the model. It seems to me that this is not indentified. You can check rank conditions. –  DreamLighter May 22 '13 at 11:13

1 Answer 1

up vote 2 down vote accepted

I think your problem is that the $t$ in the first equation should be fixed, whereas the $t$ in the second and third is actually your parameter.

The tangent line to $(t^2,t^3)$ at the point $s$ is indeed given by $(s^2,s^3)+\lambda(2s,3s^2)$ for $\lambda\in\mathbb{R}$.

This line intersects $(t^2,t^3)$ whenever $t^2 = s^2+2\lambda s$ and $t^3=s^3+3\lambda s^2$. First we can eliminate $\lambda$ to obtain $ 2t^3-3st^2+s^3=0$. Since we already know the solution $t=s$, we may factor it out: $$2t^3-3st^2+s^3 = (t-s)(2t^2-st-s^2)=0.$$

Now we may solve the resulting quadratic formula by completing the square $$2t^2-st-s^2 = 2\left(t-\frac{s}{4}\right)^2-\frac{9}{8}s^2=0.$$ Thus $t = \frac{s}{4} \pm\frac{3}{4}s$. Thus $t=s$ or $t=-\frac{1}{2}s$.

Hence, the non-trivial intersection point is at $\left(\frac{1}{4}s^2,-\frac{1}{8}s^3\right)$.

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Thank you! I think I didn't think of this approach (vectors to me) since I used calculus to get the gradient and equation of the tangent. Definitely something to keep in mind next time! –  Jerry May 22 '13 at 11:41

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