Finding point $P(a,b)$ on parabola $x=4y^2$ whose distance from the point $Q(0,33)$ is minimum and also find that minimum distance
What I try : Let coordinate of point $P$ be $(4b^2,b)$ because point $P(a,b)$ lies on $x=4y^2$
So we have $\displaystyle PQ=\sqrt{(4b^2)^2+(b-33)^2}\geq \frac{1}{\sqrt{2}}\bigg(4b^2+b-33\bigg)=\frac{4}{\sqrt{2}}\bigg[\bigg(b+\frac{1}{8}\bigg)^2-\frac{527}{64}\bigg]\geq -\frac{527}{4\sqrt{2}}$
i.e minimum occur when $\displaystyle b=-\frac{1}{8}$
But i did not understand why minimum value is negative.
Please have a look on that problem, Thanks
Note : Above we have used Inequality
$$\frac{a^2+b^2}{2}\geq \bigg(\frac{a+b}{2}\bigg)^2$$
Best Answer
The value of $b$ is the same if we are searching the minimum of $$P(b)=16b^4+b^2-66b+1089$$
The derivaitive, respect to $b$ is $$P'(b)=64b^3+2b-66$$ And the derivative of this is $$192b^2+2$$ which has no zeros and it is positive, meaning that $P'$ has only one root, and it is not zero, and that $P$ is convex everywhere. So $P(b)$ has only one minimum.
Since $P'(1)=0$, the minimum of $P$ is $P(1)=1040$.