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.

Question

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.

Answer 1

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: $$ P_{\triangle ABV} = 100\sqrt{10}.$$

Answer 2

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 $\triangle ABC$.

Answer 3

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.