What is the minimum value of $(x_1-x_2)^2+(12+\sqrt{1-x_1^2}-\sqrt{4x_2})^2$?

Question

Question: What is $$\min \left[(x_1-x_2)^2+\left(12+\sqrt{1-x_1^2}-\sqrt{4x_2}\right)^2 \right] $$ where $x_1,x_2 \in \mathbb R .$

I thought that let $x_1=x$ and $x_2=y$ and then simply minimise the function. I got $$\frac{dy}{dx}=\frac{y+12(1-x^2)^{\frac{-1}{2}}}{y+2-x+12(y)^{\frac{-1}{2}}}$$

Which in putting zero gave me $$y_o^2=\frac{144}{1-x_o^2}$$ Now how to proceed further? $(x_o,y_o)$:Critical point's coordinate

Note: This question is under conics sections heading. Am I missing something obvious form conics basics or is it completely calculus approach?

Given answer is $4\sqrt5-1$

Answer 1

It's easily shown that all of the choices are wrong . . .

Let $f$ be defined by

$$f(x_1,x_2) = (x_1-x_2)^2+\left(12+\sqrt{1-x_1^2}-\sqrt{4x_2}\right)^2$$

Then the domain of $f$ is the set of all pairs $(x_1,x_2) \in \mathbb{R}^2$ such that

$$-1 \le x_1 \le 1$$

$$x_2 \ge 0$$

Note that all of the choices for the minimum value of $f$ are less than $10$.

Suppose $f(x_1,x_2) < 10,\,$ for some $x_1,x_2$ in the domain of $f$.$\\[8pt]$

\begin{align*} \text{Then}\;\,&f(x_1,x_2) < 10\\[4pt] \implies\; &(x_1-x_2)^2+\left(12+\sqrt{1-x_1^2}-\sqrt{4x_2}\right)^2 < 10\\[4pt] \implies\; &\left(12+\sqrt{1-x_1^2}-\sqrt{4x_2}\right)^2 < 10\\[4pt] \implies\; &\left|12+\sqrt{1-x_1^2}-\sqrt{4x_2}\right|< \sqrt{10}\\[4pt] \implies\; &\left|12+\sqrt{1-x_1^2}-\sqrt{4x_2}\right|< 4\\[4pt] \implies\; &-4 < 12+\sqrt{1-x_1^2}-\sqrt{4x_2}< 4\\[4pt] \implies\; &-16 - \sqrt{1-x_1^2} < -\sqrt{4x_2}< -8 - \sqrt{1-x_1^2}\\[4pt] \implies\; &-17 < -\sqrt{4x_2}< -8\\[4pt] \implies\; &8 < \sqrt{4x_2} < 17\\[4pt] \implies\; &64 < 4x_2 < 289\\[4pt] \implies\; &x_2 > 16\\[4pt] \implies\; &|x_1-x_2| > 15\\[4pt] \implies\; &(x_1 - x_2)^2 > 15^2\\[4pt] \implies\; &f(x_1,x_2) > 15^2\\[4pt] \end{align*}

contradiction.$\\[8pt]$

Assuming I computed it correctly, the actual minimum value of $f$ is the real root $v$ of

$$v^3-433v^2+65088v-2822400 = 0$$

which yields $v \approx 72.41145380$.

The minimum is realized when $(x_1,x_2) = (1,u),\,$ where $u$ is the real root of

$$u(u+1)^2 = 144$$

which yields $u \approx 4.596909378$.

Answer 2

Wolfram Alpha also thinks the source is in error. When telling it to minimize $f(x,y)=(x - y)^2 + (12 + \sqrt{1 - x^2} - 2 \sqrt{y})^2$, it returns the function value of $\frac{1}{3}(433 - \frac{7775}{\sqrt[3]{(140112\sqrt{2919}} - 7538831)} + \sqrt[3]{140112\sqrt{2919} - 7538831})$, which does not seem to be the answer they desired.