find the equation if this tangent and the coordinates of the point where it meets the curve again.

\(\displaystyle \frac{dx}{dt} = 6t \)

\(\displaystyle \frac{dy}{dt} = 12t^{2} \)

\(\displaystyle \frac{dy}{dx} = \frac{12t^{2}}{6t} = 2t \)

when t = 1

\(\displaystyle x = 3 , y = 4 , \frac{dy}{dx} = 2 \)

equation of tangent = \(\displaystyle y -4 = 2(x-3) \)

\(\displaystyle y = 2x-2 \)

however to find the coordinates of where the curve crosses the tangent again, would I not just be able to sub in the values of x and y into the tnagent equation?

\(\displaystyle 4t^{3} = 2(3t^{2})-2 \)

\(\displaystyle 4t^{3} = 6t^{2} - 2 \) can't seem to solve from here,

any help appreicated.

thanks