Independent Variable Inequality

The given expression is nothing but the square of the distance between any two points on the quarter circle $x ^{2}+y^{2}=\theta$ and the hyperrbola $xy=16$ with both curves restricted to the first quadrant

Since it's minimum value is given as $8$ ,by this we can conclude that both curves are non intersecting

Knowing that the least distance lies along the common normal we can assume the the common normal as $y=mx$

Now by equating $-(\dfrac{dy}{dx})^{-1}=m$ from the equation of hyperbola we get $m=1 \Rightarrow y=x$ . This gives

$\Rightarrow a=\sqrt{\dfrac{\theta}{2}} \ , \ b=4$

This gives $\theta = 8 \& 72$ but at $\theta=72$ the circle will intersect the hyperbola

That's why the answer is $\large 8$

Another way of solving the question is to let $a=\sqrt{\theta}\cos \phi$ , so that the expression becomes $\theta-2\sqrt{\theta}(b\cos\phi+16/b \sin \phi)+b^2+(16/b)^2$ The above expression, for fixed $b$ achieves the minimum value, when the second term is minimum, which gives $\left(\sqrt{b^2+(16/b)^2}-\sqrt{\theta}\right)^2$ Minimizing the above with respect to $\theta$ with $\theta\ge 0$ gives $(4\sqrt{2}-\sqrt{\theta})^2$ , which is equal to $8$ . Solving the equation results in the answer $\theta=\boxed{8}$ .