Calculus 8th Edition

$x^2=-12(y-3)$
An equation of parabola when the distance of the point $(x,y)$ from the line $y=6$ is $|y-6|$ Thus,$|y-6|=\sqrt {x^2+y^{2}}$ so, $x^2=-12y+36=-12(y-3)$ $|y-6|=\sqrt {x^2+y^{2}}$ Hence, $x^2=-12(y-3)$