The parametric equations of the parabola \(y^2-8 x-4 y-12=0\) are
The parametric equations of the parabola \(y^2-8 x-4 y-12=0\) are
\(x=2+2 t^2, y=-2+4 t\)
\(x=2+4 t, y=-2+2 t^2\)
\(x=-2+2 t^2, y=2+4 t\)
\(x=-2+4 t, y=2+2 t^2\)
Solution
Given equation of parabola is
\(\begin{aligned}
y^2-8 x-4 y-12 & =0 \\
y^2-4 y & =8 x+12 \\
(y-2)^2 & =8(x+2) \\
4 a & =8 \\
a \quad a & =2
\end{aligned}\)
\(\therefore\) Parametric equations are
\(\begin{aligned}
x+2 & =a t^2, y-2=2 a t \\
x & =-2+2 t^2, y=2+4 t
\end{aligned}\)
\(\therefore\) Hence answer is (c).