A parabola has focus F and vertex V, where VF=10. Let AB be a chord of length 100 that passes through F. Determine the area of triangle VAB. 
A parabola has focus $F$ and vertex $V$, where $VF = 10$. Let $AB$ be a chord of length $100$ that passes through $F$. Determine the area of triangle $V\!AB$.

This is an olympiad question which I came across last week. I really don't have any idea where to start. I think the information provided in the question is not even enough to solve the problem.
I only know that for a parabola $y^2 = 4ax$, the length of the focal chord through $t$ is given by $a\left(t+\dfrac1t\right)^2$.
Can anyone check the problem if it's correct? If yes, then how may I proceed? Any hint would be enough.
 A: https://en.wikipedia.org/wiki/Parabola
The polar equation of parabola is $$r=\frac p{1-\cos\varphi} \tag 1 \label 1$$ where

*

*$p$ – a semi-latus rectum, which is a focal length doubled, $p=2f$,

*$r$ is a distance measured form the focus point $F$,

*$\varphi$ is a direction measured at $F$ from the axis of symmetry ($\varphi = 0$ is a direction towards the opening of parabola).

The focal length in turn is the distance from the focus of a parabola to its vertex, and it is given as $f = VF = 10.$
The endpoints of a chord, which is rotated by $\varphi$ from the parabola's axis, are at the distances given by the parabola equation $\eqref 1$:
$$\begin{cases}
r_1 = FA = \frac p{1-\cos\varphi} \\
r_2 = FB = \frac p{1-\cos(\varphi+\pi)} = \frac p{1+\cos\varphi}
\end{cases}$$
Now, the length of the chord AB, a base of our triangle $\triangle ABV$, is:
$$AB = r_1+r_2 = \frac p{1-\cos\varphi} + \frac p{1+\cos\varphi} \\
= p\,\frac{(1+\cos\varphi)+(1-\cos\varphi)}{(1-\cos\varphi)(1+\cos\varphi)} \\
= \frac{2p}{1-\cos^2\varphi} = \frac{4f}{\sin^2\varphi} \tag 2 \label 2$$
On the other hand, the height of the triangle, i.e. the distance of the vertex V from the line AB, is:
$$h = VF\,\sin\varphi = f\sin\varphi$$
From $\eqref 2$ we get:
$$\sin\varphi = \sqrt{\sin^2\varphi} = \sqrt{\frac{4f}{AB}}$$
so the area of the triangle
$$\frac 12 h\cdot AB = \frac 12 f\sin\varphi\cdot AB = \frac 12 f\cdot AB\cdot\sqrt{\frac{4f}{AB}} $$
$$\boxed{ P_{\triangle ABV} = f\cdot\sqrt{f\cdot AB}}$$
Given $f=10$ and $AB=100$ we get: $$ P_{\triangle ABV} = 100\sqrt{10}.$$
A: Consider the parabola
$ 4 p y = x^2 $
where $p = 10$
From $(0, p)$ draw a line segment making an angle $\theta$ with the horizontal direction.  The parametric equation of the line is
$ q(t) = (0, p) + s (\cos \theta , \sin \theta ) $
Intersect this line with the parabola
$ 4 p (p + s \sin \theta ) = s^2 \cos^2 \theta $
The values of $s$ that are solutions to this quadratic equation are
$ s = \dfrac{1}{2 \cos^2 \theta } \left( 4 p \sin \theta \pm \sqrt{ 16 p^2 \sin^2 \theta + 16 p^2 \cos^2 \theta} \right) = \dfrac{1}{2 \cos^2 \theta } ( 4 p \sin \theta \pm 4 p)$
The difference between the two values of $s$ is the length of the line segment, and it is equal to
$ \Delta s = \dfrac{4 p}{\cos^2 \theta} $
Set this equal to $100$ and solve for $\cos^2 \theta$
$\cos^2 \theta = 0.4 $
Now you can find the two end points of the line segment $AB$, and then calculate the area of $\triangle ABC$.
A: 
Hints:
A and B are on a circle center O on midpoint of AB. You have to find the coordinates of O somehow.Let these coordinates be $x_o, y_o$, then solve following system of equations:
$\begin{cases}y^2=40 x\\ (x-x_o)^2+(y-y_o)^2=50^2\end{cases}$
From drawing we have:
$\begin {cases}(x_A, y_A)= (78,7, 56.1)\\(x_B, y_B)=(1.27, -7.1)\end {cases}$
Which gives the coordinats of O: $(x_o, y_o)=(40,  
 24.5)$.
Now you have coordinates of $V(0. 0)$, A and B. Find measure of $VA=b$ and $VB=a$ ; you have $AB=v=100$, use Herons formula and find the area of triangle VAB.
I think the coordinates of O must be given.
