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?
 A: I come up with a solution which is based on the clue I posted above.
\begin{align*}
\frac{2}{1-a}+\frac{75}{10-b}&=\frac{2}{1-a}+\frac{15}{2}\cdot\frac{b}{10-b}+\frac{15}{2}\\
&=\frac{2\cdot \frac{a}{2}\cdot \frac{a}{2}}{(1-a)\cdot \frac{a}{2}\cdot\frac{a}{2}}+\frac{15}{2}\cdot\frac{b\cdot b}{(10-b)\cdot b}+\frac{15}{2}\\
&\ge\frac{\frac{a^2}{2}}{\left(\frac{(1-a)+\frac{a}{2}+\frac{a}{2}}{3}\right)^3}+\frac{15}{2}\cdot\frac{b^2}{\left(\frac{(10-b)+ b}{2}\right)^2}+\frac{15}{2}\\
&\ge \frac{27}{2}\left(a^2+\frac{b^2}{45}\right)+\frac{15}{2}\\
&=\frac{27}{2}\cdot1+\frac{15}{2}\\
&=21,
\end{align*}
which hold as equality iff $1-a=\dfrac{a}{2},10-b=b,$ namely $a=\dfrac{3}{2},b=5$.
A: The Tangent Line method helps.
Let $a=\frac{2}{3}x$ and $b=5y$.
Thus, $$4x^2+5y^2=9$$ and
$$\frac{2}{1-a}+\frac{75}{10-b}-21=3\left(\frac{2}{3-2x}+\frac{5}{2-y}-7\right)=$$
$$=3\left(\frac{2}{3-2x}-2-2(x^2-1)+\frac{5}{2-y}-5-\frac{5}{2}(y^2-1)\right)=$$
$$=\frac{6(x-1)^2(2x+1)}{3-2x}+\frac{15y(y-1)^2}{2(2-y)}\geq0.$$
The equality occurs for $x=y=1,$ which says that $21$ is a minimal value.
There is the following reasoning.
Let $$\frac{2}{1-a}+\frac{75}{10-b}=k.$$
Thus, $$b=10-\frac{75}{k-\frac{2}{1-a}}$$ and find all values of $k$ for which the equation
$$a^2+\frac{\left(10-\frac{75}{k-\frac{2}{1-a}}\right)^2}{45}=1$$ has solutions $0<a<1.$
A: 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 $f(0.841) = 21$.
