Question

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

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

Answer 1

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

Given: $x^2-4x-32=-12y$
$\Rightarrow x^2-4x+4-4-32=-12y$
$\Rightarrow (x-2{)}^2=-12(y-3)$

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 $X=x-2,Y=y-3$, we get
$\left(i\right)x-2=2at\Rightarrow x=2+6t\cdots \left(1\right)\left(ii\right)y-3=-a{t}^2\Rightarrow y=3-3t^2\cdots \left(2\right)$

Hence required parametric form:
$x=2+6t,y=3-3t^2$