Parametric Curves - Tangent to the curve

Question

I am struggling with finding the tangents to a curve below that passes through point (12,9). $$x=9t^2 +3 ,\qquad y=6t^3 +3$$

I have $dx/dt= 18t,\enspace dy/dt=18t^2$, which then gives us $0= -3t^2+15t-6$ Now, I think this is where my problem is coming in: I pulled out -3 to get $0= -3(t^2-5t+2)$ which is not easily/rationally factorable.

So the equations I came up with (there should be 2) are: $y=x-21$ and $y=-2x+24$

Can anyone point me in the correct direction/assist me in determining the answers.

Answer 1

Since the point (12,9) is on the curve, we can determine that $t=1$. Calculate the slope of this point,

$$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}= \frac{18t^2}{18t} = 1$$

Then, use the point-slope equation to obtain the tangent line $y-9=x-12$, or,

$$y = x -3$$