Question

Find the parametric equation of the parabola $x^2-4x-32=-12y$

• A. $x=3+6t,y=2-3t^2$
• B. $x=2-6t,y=3+3t^2$
• C. $x=6t,y=3t^2$
• D. $x=2+6t,y=3-3t^2$

Answer 1

The correct option is D $x=2+6t,y=3-3t^2$

$x^2-4x-32=-12y$
$\Rightarrow x^2-4x+4-4-32=-12y$
$\Rightarrow (x-2{)}^2=-12(y-3)$
Comparing the given equation with $X^2=-4aY$
$a=3,X=x-2,Y=y-3$

Parametric form for $X^2=-4aY$ is
$X=2at,Y=-a{t}^2$
Put $X=x-2,Y=y-3$
$x-2=2at\Rightarrow x=2+2at\Rightarrow x=2+6ty-3=-a{t}^2\Rightarrow y=3-a{t}^2\Rightarrow y=3-3t^2$