# Thread: Isosceles Triangle in a Parabola help

1. ## Isosceles Triangle in a Parabola help

Triangle OAB is an isosceles triangle with vertex O at the origin and vertices A and B on the parabola y = 9-x^2

Express the area of the triangle as a function of the x-coordinate of A.

I'm having trouble finding the sides of all possible isosceles triangles that has its vertex in y= 9 -x^2. I know that the distance formula helps, so I can probably substitute the lengths by using A = 1/2 bh of a circle or maybe heron's formula. I will be glad if someone could help to figure out the equation for this.

2. Originally Posted by Enderless
Triangle OAB is an isosceles triangle with vertex O at the origin and vertices A and B on the parabola y = 9-x^2

Express the area of the triangle as a function of the x-coordinate of A.

...
Hello,

you have to find the main condidtion which contains the variable which should get an extreme value. With your problem it is the variable for the area: a.

1. main condition: $\displaystyle a=\frac{1}{2}\cdot b \cdot h$

2. aux. conditions:
i) If the point A has the coordinates A(p, 9-pē), p > 0, then the base b of the triangle is: $\displaystyle b = 2 \cdot p$
ii) then the height h is: $\displaystyle h = 9-p^2$

3. Substitute the variables from the aux. condition into the equation of the main condition and you'll get the characteristic function:
$\displaystyle a(p)=\frac{1}{2}\cdot 2 \cdot p \cdot (9-p^2)=-p^3+9p$

4. Derivate a wrt p: $\displaystyle a'(p) = -3p^2+9$. To get the extreme value a'(p) = 0. Solve for p.

5. You'll get $\displaystyle p = \sqrt{3}$. The point A has the coordinates A($\displaystyle \sqrt{3}$, 6) and the maximum value of the area is:

$\displaystyle a(\sqrt{3}) = -(\sqrt{3})^3+9\cdot \sqrt{3} = 6\cdot \sqrt{3} \approx 10.39~ units^2$

6. I've attached a sketch of the situation.

3. Thank you for the reply, the equation works, but only for a certain number of triangles embedded in the parabola. I'm trying to find an equation that determines the area of all possible isosceles triangles.

But what if points A and B do not share the same x-coordinates?

Here's an example, O = (0,0)
A = (2,5)
B = (-2.2777902,3.8111609)
And the area will be around 4.75 units^2

The points above can still make an isosceles triangle, with a similar triangle having the points

O = (0,0)
A = (2,5)
B = (-2,5)
The area here will be 10 units^2

The equation will only work for the second triangle, but not for the first. Is there a way to find an equation that works for all possible triangles?

4. Originally Posted by Enderless
Thank you for the reply, the equation works, but only for a certain number of triangles embedded in the parabola. I'm trying to find an equation that determines the area of all possible isosceles triangles.

But what if points A and B do not share the same x-coordinates?
...
Hello,

I've attached a drawing of the situation as I understand it. If you mean this, it will be quite difficult to find the required equation:

1. Find the equation of the perpendicular bisect of OB.

2. The perp. bisect intersect the parabola in A.

3. the length of OB is the base and the length of the perp. bisect from the midpoint of OB to A is the height of the triangle.

4. All terms had to be dependent on the x-value of B (That's the point marked as a red cross in my drawing)

5. Good luck!

5. I find it very complicated to express B in terms of a in the x- coordinate when B and A have different x coordinates.

Well, thank ya though.

Does anyone else know how to solve this?

,

,

### Prove that triangle OAB is an isosceles triangle

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