Answer
(a) $y^2=12x$
(b) $31cm$
Work Step by Step
(a) Assume the equation for the parabola is $y^2=4px$, so the focus is at $x=p$. Since when $x=p, y=\pm2p$ and the focal diameter is 12cm, we have $4p=12, p=3cm$ and the equation becomes $y^2=12x$
(b) Given $x=20cm$, we have $y^2=240$ and we have $y=\pm 4\sqrt {15}$ and the diameter of the opening $d_{\bar {CD}}=8\sqrt {15}\approx31cm$
