# Finding Tangent and Intersection for Parametric Curve

1. Apr 23, 2012

### Seydlitz

1. The problem statement, all variables and given/known data
A curve is defined by the parametric equations:

x = 2t^3
y = 2t^2
t =/ 0

1)Prove that the equation of the tangent at the point with parameter t is 2x - 3ty + 2t^3 = 0. Proven, and I've no problem with this part.

2.)The tangent at the point t = 2 meets the curve again at the point where t = u. Find the value of u.

3. The attempt at a solution

I substituted the value of t = 2 to the first equation in problem 1.
It gives me 2x - 6y + 16 = 0
y = (1/6)(2x + 16)

At this point I'm stuck because ordinarily I'll just set the tangent equation equal to the curve, expressed in terms of x and then solve for x. But now with parametric equation I do not know how to proceed next or how to see the problem.

What is the condition required for a tangent to be said intersecting a parametric curve?

Thank You

2. Apr 23, 2012

### Seydlitz

Update: After a long moment of gedankenexperiment, :D

I've decided to substitute the initial curve parameter to the equation of the tangent with t = 2. That is 2x - 6y + 16 = 0. x = 2t^3, y = 2t^2. I already know one of the root and proceed to factorize the equation. Which gives me the solution, t = 2 and t = -1. So u = -1.

3. Apr 23, 2012

### sharks

At t=u, find the parametric equations for x and y. Then replace in your equation of tangent for t=2.

But what you've done is correct as well. The solution for "u" might not have been so obvious if the question was not simply t=u, for example, t=(2u-4).

4. Apr 23, 2012

### Seydlitz

Is my solution acceptable? (2nd post)

5. Apr 23, 2012

### Whovian

Remember. $\displaystyle\dfrac{\mathrm{d}y}{\mathrm{d}x}= \dfrac{\frac{\mathrm{d}y}{\mathrm{d}t}}{\frac{ \mathrm{d}x}{\mathrm{d}t}}$. It turns out that the d's are actual terms, but most analysis courses don't teach that and tell us to realize this isn't "proper" math because understanding them is enormously difficult. So always keep in mind that you _can_ treat them like fractions. Just don't tell your teacher. :)

6. Apr 23, 2012

### sharks

If the 2nd point of intersection is (A,B), then if you put x=A into the equation of tangent, making y the subject of formula, you should get the answer B (that is, the y-coordinate of the 2nd point).

7. Apr 23, 2012

### Seydlitz

Ahh that's the case! I've been doing it a bit unconsciously with parametric problem then suddenly I know there's deep secret underneath it. Pretty neat.

Alright, I've checked my solution accordingly. Thank You!