Area of isosceles triangle

Question

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 ?

Answer 1

Assume $B(t^2,2t)$ and $C(t^2,-2t)$ as $AB=AC$, Where $t>0$.

As you said orthocentre is $H(0,0)$, it implies that,

$m_{BH}\cdot m_{AC}=-1$

Where $m_l$ represents slope of line $l$.

Or,

$$\left(\frac{2t-0}{t^2-0}\right)\left(\frac{-2t-0}{t^2+1}\right)=-1$$$$\implies t=\sqrt{3}$$

As $t>0$

So, area will be $$ \frac{1}{2} (1+t^2)(4t)=\boxed{8\sqrt{3}}$$

$\frac{1}{2}×$distance from $A$ to $BC$$×BC$

Answer 2

Since another more algebraic answer was posted, I see no reason not to also post my original attempt. It requires some more complex geometrical concepts, but also a bit less algebra.

Suppose $B$ has coordinates $(x, y).$ We would then have $\vec{AB} = \langle x + 1, y\rangle,$ since we're going from $(-1, 0)$ to $(x, y).$

Note that by symmetry we have that $\vec{C} = \langle x, -y\rangle,$ and that because the origin is the orthocenter of the triangle, $\vec{C}$ is orthogonal to $\vec{AB},$ so

$$\vec{C} \cdot \vec{AB} = \langle x + 1, y\rangle \cdot \langle x, -y\rangle = (x + 1)(x) + (y)(-y) = x^2 + x - y^2 = 0$$

Now recalling that our restriction on $B$ requires that $y^2 = 4x,$ we have

$$x^2 + x - 4x = 0 \Rightarrow x^2 - 3x = 0 \Rightarrow x = 0 \text{ or } x = 3$$

and because $x = 0$ would give us a degenerate triangle, we must have $x = 3.$ From here we can proceed the same way as the other answer, showing that our triangle has altitude $4$ and base $4\sqrt{3},$ so its area is $$\frac12 \cdot 4 \cdot 4\sqrt{3} = \boxed{8\sqrt{3}}$$

Answer 3

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 $2 \sqrt{3} (3 + 1) = \boxed{8 \sqrt3}$.