Parametrize intersection of surface and plane

Question

Parametrize the intersection of the surface $x^2 + y + z = 3$ and the plane $-2x + 4y - 3z + 1 = 0$.

I eliminated the y variable and found the intersection of the two surfaces to be $(x+\frac14)^2+\frac{7z}4 = \frac{53}{16}$. But what do I do next? I know I have to do substitute a variable with $t$, but I'm stuck.

Answer 1

$(x+\frac 14)^2+\frac {7z}{4}=\frac{53}{16}$

say: $x = t- \frac 14$

$t^2+\frac {7z}{4}=\frac{53}{16}\\ 7z =\frac{53}{4} - 4t^2\\ z =\frac{53}{28} - \frac 47t^2$

now plug into one of your original equations

$−2t + \frac 12 +4y−\frac {159}{28}+\frac {12}7 t^2 +1=0$

And solve for $y$