Finding points on a tangent line of a parametric equation that are parallel to another parametric equation

Question

So I got the tangent line of the first equation to be 12t/(3t^2+4) and I changed the second parametric equation to the cartesian form and got y= -(12/7x+5) with 12/7 as my slope. I equated 12t/(3t^2+4)= 12/7 and solve for t using the quadratic formula. I got the values -1 and -4/3 but they are incorrect. So, I'm confused where I went wrong.

Answer 1

From $y = (-12/7)x - 5$, the slope of the second line should be -12/7 instead.

Also, the solutions to $\frac{12t}{3t^2+4} = \frac{12}{7}$ are $1, \frac{4}{3}$ and $-1, -\frac{4}{3}$ are instead the solutions to $\frac{12t}{3t^2+4} = -\frac{12}{7}$

For the points $(y(t), x(t))$ of the solution, you should plug in $t = -1, -4/3$ into your equation for $x,y$.

Otherwise , your general approach is fine.