shortest distance from moving point in parabola to point

Question

P is the moving point on the following parabola and Q is fixed point (-1,8) .calculate shortest distance between p and q

$2x=y^2$

$x=\frac{y^2}{2}$

$d= \sqrt{(y-8)^2 + (x+1)^2}$

$d^2= (y-8)^2 + (\frac{y^2}{2}+1)^2$

$d^2= 2y^2-16y+\frac{y^4}{4}+65$

$d`=4y-16+y^3=0$

$y=16$ or $y=\sqrt{12}$

$y^2=2x$

$16^2=2x$

$x=128$

$y^2+4=16$

$(128,16)$

$y^2=12$

$y=\sqrt{12}$

$(6,\sqrt{12})$

is this right?

Answer 1

No, it's wrong.

The equation $y^3+4y-16=0$ gives $y=2$.

My solution.

By your work and by AM-GM $$d=\sqrt{(y-8)^2+\left(\frac{y^2}{2}+1\right)^2}=\sqrt{\frac{y^4}{4}+2y^2-16y+65}=$$ $$=\sqrt{\frac{y^4}{4}+2\cdot y^2+5\cdot4-16y+45}\geq\sqrt{8\sqrt[8]{\frac{y^4}{4}\cdot(y^2)^2\cdot4^5}-16y+45}=$$ $$=\sqrt{16|y|-16y+45}\geq\sqrt{45}=3\sqrt5.$$ The equality occurs for $y=2$, which says that $3\sqrt5$ is a minimal value.