Find the equations of the tangents to the given curve that pass through the point (15, 15).

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

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

I isolated t from x:

$\displaystyle t=\pm \sqrt {\frac {x-6}{9}}$

Plugged it into the y equation:

$\displaystyle y = \pm 6\left(\sqrt {\frac {x-6}{9}}\right)^3 + 9$

$\displaystyle y = \pm 162\left(\sqrt {x-6}\right)^3 + 9$

$\displaystyle y = \pm 162(x-6)^{\frac {3}{2}} + 9$

Then found the the derivatives of y:

$\displaystyle y'=\pm 243\sqrt {x-6}$

Plugged in when x and y are equal to 15:

$\displaystyle 15=\pm 243\sqrt {15-6}\cdot 15 +b$

$\displaystyle

b = 10950, 10920$

So I found my two equation to be:

$\displaystyle

y=243\sqrt {x-6} + 10920$

and

$\displaystyle y=-243\sqrt {x-6} + 10950$

But these aren't the right answers. Any suggestions?