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


Combining the results in the previous discussion and answers, we set $x=a t$, then $y=b\sqrt{1-t^2}$ and $dx=adt$.  So:
$$\bar{x}=\frac{\int_0^a xy(x)\,\mathrm{d}x}{\int_0^a y(x)\,\mathrm{d}x}=\frac{a^2b}{ab}\frac{\int_0^1 t \sqrt{1-t^2}\,\mathrm{d}t}{\int_0^1 \sqrt{1-t^2}\,\mathrm{d}t}=a\frac{1/3}{\pi/4}=\frac{4a}{3\pi}$$
Similarly we obtain:
$$\bar{y}=\frac{4b}{3\pi}$$
Thus
$$\bar{x}=\frac{4a}{3\pi}=4 \implies a=3\pi$$
$$\bar{y}=\frac{4b}{3\pi}=2 \implies b=\frac{3\pi}{2}$$
