# Thread: Tangent to a parametric curve

1. ## Tangent to a parametric curve

I thought I solved this correctly, but my answer was marked wrong:

Here it is: Find the equations of the tangents to this curve that pass through the point (15,15).

$\displaystyle x=9t^2+6$
$\displaystyle y=6t^3+9$

Solution:
$\displaystyle \frac{dy}{dx}=\frac{18t^2}{18t} = t$

$\displaystyle y-(6t^3+9)=t(x-(9t^2+6))$

Substituting in the point (15,15):

$\displaystyle 15-6t^3-9=t(15-9t^2-6)$

After simplifying, this turns into: $\displaystyle (t-1)^2(t+2) = 0$

Setting t=1:
$\displaystyle y-15=1(x-15)$
$\displaystyle y=x$ is my first tangent line, which was wrong. When I graphed it, it did appear to be tangent to the curve at that point..

Setting t=-2:
$\displaystyle y-15=-2(x-15)$
$\displaystyle y=-2x+45$ also incorrect.

If someone could fix my work, I'd be very very thankful.

2. Both those answers look correct to me. Why do you say that they are wrong?

3. t= -2 is incorrect because y(-2)= 6(-2)^3+ 9= -48+ 9= -39, not 15.

However, at t= 1, (x, y)= (15, 15) and then tangent line is indeed y= x.

4. Originally Posted by HallsofIvy
t= -2 is incorrect because y(-2)= 6(-2)^3+ 9= -48+ 9= -39, not 15.
I thought the same at first. But when t=–2, x=42 and y=–39, and the tangent at (42,–39) is y+39 = –2(x–42) which does indeed simplify to $\displaystyle y = -2x+45$.