<s>[INST] Area of isosceles triangle I need to solve following problem:

The vertex of the isosceles triangle $ABC$ is the point $A(-1, 0)$, and the vertices $B$ and $C$ belong to the parabola $y^2 = 4x$. If the point $(0, 0)$ is the orthocenter of triangle $ABC$, then its area is equal to ?

I've tried to use fact that points C and B lies on parabola to express their coordinates $B(x,2\sqrt x)$ and $C(x,-2\sqrt x)$
Also i can express height and area of triangle as:
Area = $2\sqrt{x}(x+1)$.
Height = $x+1$
Can you help me to figure it out ?
 [/INST] Given that the line connecting $B$ to the origin is $y = mx$, $y^2 = 4x$ gives $m^2 x^2 - 4x = 0$ $ \implies x(m^2 x - 4) = 0$, so the other intersections apart from $(0, 0)$ are when $x = \frac{4}{m^2}$, and $y = \pm \frac{4}{m}$.
Now finding the equation of the perpendicular which must pass through $(-1, 0)$, we have:
$$y - 0 = -\frac{1}{m} (x - -1)$$
$$-\frac{4}{m} = -\frac{1}{m} (\frac{4}{m^2} + 1)$$
$$-4m^2 = -4 - m^2$$
$$4 = 3m^2, m^2 = 4/3$$
and thus $x = 4/(4/3) = 3$ leading to an area of $2 \sqrt{3} (3 + 1) = \boxed{8 \sqrt3}$.
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