Solve $a$ and $b$ for centre of mass in $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ Given ellipse:
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
What length do $a$ and $b$ have to be so the centre of mass is $S(4;2)$?
I've tried steps to solve the equation to $$y=b\sqrt{1-\frac{x^2}{a^2}}$$ and integrate 
$$A=b\int_0^a{\sqrt{1-\frac{x^2}{a^2}}}$$
But I'm not achieving a satisfying result. There must be an easier way . Enlighten me please


If you knew the location of the center of mass (COM) for a quarter circle, it'd be easy: you'd just find the scaling-transform that mapped that point to $(4, 2)$. By symmetry, the COM for the quarter-circle must be at some point $(s, s)$ along the line $y = x$. But I cannot see any way, other than actually doing the integration, to find it. The denominator, in this case, is easy -- it's just the area of the quarter circle, i.e., it's $\pi/4$. The numerator is 
\begin{align}
\int_0^1 x \sqrt{1 - x^2} dx &= \frac{1}{2} \int_0^1 2x \sqrt{1 - x^2} dx\\
&= \frac{1}{2} \int_1^0 -\sqrt{u} du \text{, substituting $u = 1 - x^2$}\\
&= \frac{1}{2} \int_0^1 u^{1/2} du \\
&= \frac{1}{2} \left.\frac{u^{3/2}}{3/2}\right|_0^1 \\
&= \frac{1}{2} (\frac{2}{3}) \\
&= \frac{1}{3}. 
\end{align}
That makes the $x$-coord of the centroid (for a circle) be $s = \frac{1/3}{\pi/4} = \frac{4}{3\pi}$. And the centroid is at location $(s, s)$. 
What I'm wondering, and hoping other stackexchangers might be able to suggest, is a geometric argument for this result that makes it completely obvious without integration. Don't ask much, do I? 
