Minimize $\frac{2}{1-a}+\frac{75}{10-b}$ 
Let $a,b>0$ and satisfy $a^2+\dfrac{b^2}{45}=1$. Find the minimum
  value of  $\dfrac{2}{1-a}+\dfrac{75}{10-b}.$

WA gives the result that $\min\left(\dfrac{2}{1-a}+\dfrac{75}{10-b}\right)=21$ with $a=\dfrac{2}{3},b=5$.
Consider making a transformation like this
$$\dfrac{2}{1-a}+\dfrac{75}{10-b}=\frac{2\cdot (ma)^2}{ma\cdot ma\cdot (1-a)}+\frac{75\cdot(nb)^2}{nb\cdot nb\cdot(10-b)}.$$
If $2m=2n=1$, $ma=1-a$, and $ nb=10-b$ can all hold, we can apply AM-GM inequality as follows
$$\frac{2\cdot (ma)^2}{ma\cdot ma\cdot (1-a)}+\frac{75\cdot(nb)^2}{nb\cdot nb\cdot(10-b)}\geq \frac{2\cdot (ma)^2}{\left(\frac{ma
+ma+(1-a)}{3}\right)^3}+\frac{75\cdot(nb)^2}{\left(\frac{nb+nb+(10-b)}{3}\right)^3}=\cdots$$
But this is invalid, since $n=\frac{1}{2},nb=10-b$ can not satisfy $b=5$.
How to solve it?

There is another way using parameterisation. Let $a = \cos t, b = \sqrt{45} \sin t$, then we have:
$$f(t) = \frac{2}{1-\cos t} + \frac{75}{10-\sqrt{45} \sin t}$$
$$f'(t) = -\frac{2 \sin t}{(1-\cos t)^2} + \frac{225 \sqrt5 \cos t}{(10-3 \sqrt{5} \sin t)^2}$$
The condition that $a,b > 0$ translates to $0 < t < \frac{\pi}{2}$. Setting $f'(t)$ equal to $0$, we find that only $0.841$ (actually $\tan^{-1} \left( \frac{1}{\sqrt5} \right)$) is inside the range. 
Therefore the minimum value is $f(0.841) = 21$.
