Ellipse and chord length There is a analytic geometry problem:

In the ellipse $\frac{x^2}{4}+y^2=1$, segment $AB$ is a chord and $AB=\sqrt{3}$, find the maximum and minimum area of $\triangle AOB$.


My progress
Assume $A(x_1,y_1),B(x_2,y_2)$, we have
$$ \left\{ 
\begin{array}{}
(x_1-x_2)^2+(y_1-y_2)^2=3 \\
\frac{x_1^2}{4}+y_1^2=1 \\
\frac{x_2^2}{4}+y_2^2=1
\end{array}
\right.$$
And
$$ S_{\triangle AOB}=\frac{1}{2}(x_1y_2-x_2y_1)$$
After a complicated simplify, I got
$$ \frac{3}{4}x_1^2+\frac{3}{4}x_2^2-2x_1x_2-\frac{\sqrt{16-4x_1^2-4x_2^2+x_1^2x_2^2}}{2}=1$$
and the target $\frac{1}{2}(x_1y_2-x_2y_1)$ turned to
$$ \frac{\sqrt{4x_1^2-x_1^2x_2^2}-\sqrt{4x_2^2-x_1^2x_2^2}}{2}$$
It likely is a conditional extremum problem and I don't know what to do next.
 A: 
It likely is a conditional extremum problem and I don't know what to do next.

I don't know what to do next either.
So, let us take another approach.
If the line $AB$ is parallel to $y$-axis, we can easily see that the area of the triangle is $\sqrt 3/2$.
If the line $AB$ is not parallel to $y$-axis, we can set $AB : y=ax+b$.
Then, 
$$\frac{x^2}{4}+(ax+b)^2=1\iff (4a^2+1)x^2+8abx+4b^2-4=0$$
Let $\alpha,\beta$ be the $x$ coordinates of $A,B$. Then,
$$\alpha+\beta=\frac{-8ab}{4a^2+1},\quad\alpha\beta=\frac{4b^2-4}{4a^2+1}$$
Now, we can write
$$\sqrt 3=AB=\sqrt{(\alpha-\beta)^2+((a\alpha+b)-(a\beta+b))^2}$$
Squaring the both sides gives
$$3={(a^2+1)((\alpha+\beta)^2-4\alpha\beta)}={(a^2+1)\left(\left(\frac{-8ab}{4a^2+1}\right)^2-4\frac{4b^2-4}{4a^2+1}\right)}$$
and so 
$$b^2=\frac{16a^4+56a^2+13}{16(a^2+1)}$$
Let $S$ be the area of the triangle. Then, 
$$S^2=\left(\frac{\sqrt 3}{2}\cdot\frac{|a\cdot 0-0+b|}{\sqrt{a^2+1}}\right)^2=\frac{3b^2}{4(a^2+1)}=\frac{3(16a^4+56a^2+13)}{64(a^2+1)^2}:=f(a)$$
$$f'(a)=\frac{9a(5-4a^2)}{16(a^2+1)^3}=0\iff a=0,\pm\frac{\sqrt 5}{2}$$
Hence, the maximum of $f(a)$ is $f(\pm\sqrt 5/2)=1$, and the minimum of $f(a)$ is $f(0)=39/64$.
Therefore, the maximum area of the triangle is $\color{red}{1}$, and the minimum area is $\color{red}{\sqrt{39}/8}$.
