Finding the point on the parabola closest to a point

Question

Find the point on the parabola $ 2x - 2y^2 = 7$ which is closest to the point $ (4,16) $.

I've tried the distance $ D $ between the point $ (x,y) $ and $ (4,16) $, then the problem is simplified by considering $ D^2 $ instead of $ D $ as the function. It is also easier to get the idea on the bounds of $ x $ or $ y $ if you sketched the graph.

Answer 1

$$D=\sqrt{(x-4)^2+(y-16)^2}=$$ $$=\sqrt{(y^2+3.5-4)^2+(y-16)^2}$$ $$D^2(y)=y^4-32 y+256.25 \ge \frac{833}4 \left(D'=4y^3-32=0 \Leftrightarrow y=2\right)$$ at $y=2$

Then $(x;y)=(\frac{15}2;2)$

Answer 2

Call $f(x)=d^2={(x-4)^2+(\sqrt{x-7/2}-16)^2}$

And then look for the roots of $f'(x)$.