Triangle of maximal area with vertices origin and intersections of a chord with a curve "let lx + my=1 be a chord  of the curve, $3x^2 + y^2 - 2x+4y=0$, interesecting the curve at points A and B such that AB subtends a right angle at origin 'O'. If the triangle OAB is isoceles then the area of triangle can not exceed ___?"
My attempts:
Attemp#1:
So if I understood right, chord is basically a secant on the curve. Since the curve is a quadratic implicit function, I'm conjecturing the secant line they give is either a line parallel to x axis or y axis. However if you start actually working on this, it'll turn very ugly ( you can definitely find 'm' and 'x' values tho)
Attempt #2:
I tried to optimize the function in terms of 'y' , and using that height as the base of my triangle as the optimum 'y' and multiplying it by x coordinate of the optimum 'y' and halfing for max area
edit: options are 5,6,7,8
 A: First of all we need to find the points of intersection of two lines, i.e. solve the system:
$$
\begin{align}
3x^2-2x+y^2+4y=0\tag1\\
mx+ny=1\tag2
\end{align}
$$
Substituting $y=\frac{1-mx}n$ from (2) into (1) one ends up with a quadratic equation for $x$ which has 2 real roots $x_1, x_2$ provided that the discriminant of the equation satisfies:
$$
4m^2+n^2+4mn+2m-12n-3>0.
$$
The coordinates $y_1,y_2$ can be then found from (2).
The area of the triangle is
$$
\frac{|x_1y_2-x_2y_1|}2\stackrel{!}=\frac{\sqrt{4m^2+n^2+4mn+2m-12n-3}}{m^2+3n^2}\tag3
$$
where some boring but straightforward algebra is hidden behind the relation $\stackrel{!}=$.
The angle subtended by the points $(x_1,y_1)$ and $(x_2,y_2)$ is right if and only if
$$
x_1^2+y_1^2+x_2^2+y_2^2-(x_2-x_1)^2-(y_2-y_1)^2=0\stackrel{!}\implies 8-4m+8n=0.\tag4
$$
The triangle is isosceles  if and only if
$$
x_1^2+y_1^2-x_2^2-y_2^2=0\stackrel{!}\implies 2m^3+n^3+m^2n+2mn^2-2mn=0.\tag5
$$
It can be checked that the equations (4) and (5) have a single common solution, which corresponds to the real root of the equation:
$$
25n^3+56n^2+48n+16=0.
$$
Numerically $n\approx-0.891727$.
A: Hint
It's not difficult to find out 
that the given curve is in fact an ellipse
with the equation in a standard form
\begin{align}
\frac{(x-x_0)^2}{a^2}
+
\frac{(y-y_0)^2}{b^2}
&=1
,
\end{align}
its venter is located at the point
\begin{align}
C=(x_9,y_0)
&=
(\tfrac13,\,
-2)
\end{align}
the semi-axes
\begin{align}
a&=\tfrac{\sqrt{13}}3
,\quad
b=\sqrt{\tfrac{13}3}
,
\end{align}
the distance between the foci is
\begin{align} 
d&=\sqrt{b^2-a^2/4}
=\tfrac16\sqrt{143}
\approx 1.9930
,
\end{align} 
the focal points are
\begin{align} 
F_0 &= \Big(\tfrac13, -2-\tfrac1{12}\sqrt{143} \Big)
,\\
F_1 &= \Big(\tfrac13, -2+\tfrac1{12}\sqrt{143} \Big)
\end{align}

The upper function of the curve is
$f(x)=-2+\sqrt{4-3x^2+2x}$, use
variables $A_x,B_x$ in
$A=(A_x,f(A_x))$, $B=(B_x,f(B_x))$
and constraints $|A|=|B|$,
$\Re(A/B)=0$. 
I've got 
\begin{align} 
B_x&=
\tfrac16
+\tfrac16\,\sqrt[3]{433+21\sqrt{465}}
-\tfrac{13}3\,\frac1{\sqrt[3]{433+21\sqrt{465}}}
\approx 1.31613140
,\\
|OB|^2=|OA|^2&\approx
2.37510
, 
\end{align}
so the area of $\triangle AOB\approx 1.18755$
and we are safe to declare that 
the area can not exceed, say, $1.2$.
