# Thread: Help: triangle under parabola

1. ## Help: triangle under parabola

I'm not sure if this is algebra or geometry; I haven't studied math in a very long time. Anyway, I really need help with this problem. I've attached an image of the problem, and here's a translation of the text from Spanish:

An isosceles rectangular triangle is inscribed in the parabola y^2=4x with the right angle in the vertex of the curve like shows the figure. The area of the triangle AVC is: (as shown)

I've already spent about an hour tearing my hair out and looking through books to try to figure out how to do this. Any help would be greatly appreciated. Thanks a bunch

2. Originally Posted by mpoderoso
I'm not sure if this is algebra or geometry; I haven't studied math in a very long time. Anyway, I really need help with this problem. I've attached an image of the problem, and here's a translation of the text from Spanish:

An isosceles rectangular triangle is inscribed in the parabola y^2=4x with the right angle in the vertex of the curve like shows the figure. The area of the triangle AVC is: (as shown)

I've already spent about an hour tearing my hair out and looking through books to try to figure out how to do this. Any help would be greatly appreciated. Thanks a bunch
The base of the triangle is $\displaystyle 2q$ and the height is $\displaystyle p$.

The area of a triangle is $\displaystyle \frac{1}{2}\times\textrm{ base }\times\textrm{ height}$

$\displaystyle = \frac{1}{2}\times 2q\times p$

$\displaystyle = pq$.

3. Is this supposed to be a multiple choice? The image gives me that impression, but none of the answers make sense.

You can substitute p and q into the parabola's equation and end up with $\displaystyle q = \pm \sqrt{4p}$, and you can solve for p as $\displaystyle p = \frac {q^2}{4}$. As Prove It says, the area is clearly pq. We can substitute and get the area in terms of one or the other variable and end up with: $\displaystyle A = 2p^{\frac {3}{2}}$ and $\displaystyle A = \frac {q^3}{4}$.

Anyways, I tried to see which of those formulas in the image gave the correct answer, and none of them seem to. For example, with p = 9, we have q = 6, giving an area of 54. None of those formulas gives 54 as the answer.

So if that is multiple choice, either I'm confused or they are all incorrect.

4. Hello, mpoderoso!

Are you sure of the translation?
None of the answers makes sense.
In fact, none of it makes any sense.

An isosceles right triangle is inscribed in the parabola $\displaystyle y^2\,=\,4x$
with the right angle in the vertex of the curve like shows the figure.

The area of the triangle AVC is:

. . $\displaystyle (A)\;\frac{q^2+4q}{2} \qquad(B)\;p^2+2p \qquad(C)\;\frac{p^2+4p}{2} \qquad (D) \;q^2+2q$ .??

Code:
|         A
|         * (p,q)
|   *   / |
| *   /   | q
|*  /     |
| /       |
V * - - - - + - - - -
| \   p   |
|*  \     |
| *   \   | q
|   *   \ |
|         * (p,-q)
|         C
|

The base of the triangle is: $\displaystyle AC \,=\,2q$
The height of the triangle is: $\displaystyle p$

The area is: .$\displaystyle A \;=\;\tfrac{1}{2}\text{(base)(height)} \;=\;\tfrac{1}{2}(2q)(p) \quad\Rightarrow\quad\boxed{A\;=\;pq}$

Since they gave us the equation of the parabola: .$\displaystyle y^2 \,=\,4x$
. . we can locate vertices $\displaystyle A$ and $\displaystyle C$ exactly.

Sides $\displaystyle VA$ and $\displaystyle VC$ are "45°-lines".
. . Their equations are: .$\displaystyle y = x$ and $\displaystyle y = -x$
They intersect the parabola at: .(4,4) and (4,-4).

. . Therefore, the base is $\displaystyle 8$ and the height is $\displaystyle 4$ . . . The area is $\displaystyle {\color{blue}16}$.

So what's with the $\displaystyle p's$ and $\displaystyle q's$ ?