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.