Isocleles Triangles in a Parabola Help

In summary, this problem asks for an equation to find the area of an isosceles triangle in a given parabola as a function of the x coordinate of the vertex. There are several ways to solve for the area, depending on whether the "base" of the triangle is parallel to the x-axis or not.
  • #1
Enderless
4
0

Homework Statement



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.

Homework Equations



A = 1/2 bh
Distance formula (maybe)
Heron's Formula (an alternative)

The Attempt at a Solution



The area of a triangle is given by A = 1/2 bh. If point A is (x, 9 - x^2), then b=2x and h = 9-p^2
The final equation will be A = 9x - x^3 (as a function of the x coordinate of A)

The equation above works only for a limited number of triangles embedded in the parabola. For instance, if A and B has the same x coordinates, then the equation will work. But i considered two different triangles that can make an isosceles triangle:

O = (0,0)
A = (2,5)
B = (2,5)
Area is 10 unites^2

and

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

Both of those are isosceles triangles, that can fit into the parabola of 9-x^2. How would I go about finding an equation that can find the area of all possible isosceles triangles in the parabola?
 
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  • #2
Enderless said:

Homework Statement



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.


Homework Equations



A = 1/2 bh
Distance formula (maybe)
Heron's Formula (an alternative)


The Attempt at a Solution



The area of a triangle is given by A = 1/2 bh. If point A is (x, 9 - x^2), then b=2x and h = 9-p^2
You mean 9- x2, of course.

The final equation will be A = 9x - x^3 (as a function of the x coordinate of A)

The equation above works only for a limited number of triangles embedded in the parabola. For instance, if A and B has the same x coordinates, then the equation will work. But i considered two different triangles that can make an isosceles triangle:

O = (0,0)
A = (2,5)
B = (2,5)
Area is 10 unites^2

and

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

Both of those are isosceles triangles, that can fit into the parabola of 9-x^2. How would I go about finding an equation that can find the area of all possible isosceles triangles in the parabola?
Yes, if OA and OB are the two congruent sides then the problem is easy as you say. Suppose, instead, that OA and AB are the conguent sides. Let (x0,y0) be the point A (and we may, without loss of generality, assume that x0 is negative. B= (x,y) must satisfy y= 9- x2 and [itex](x-x_0)^2+ (y-y_0)^2= x^2+ y^2[/itex] If you multiply out the left side, you will see that the x2 and y2 terms cancel leaving [itex]-2x_0x+ x_0^2- 2y_0y+ y_0^2= 0[/itex] or [itex](2x_0)x+ (2y_0)y= x_0^2+ y_0^2[/itex]. It should not be too difficult to solve those two equations for x and y and then calculate the area of the triangle in terms of x0 and y0.

The case where OB and AB are the congruent sides is just the reflection of the previous case in the y-axis and gives the same area formula.
 
  • #3
Another way is, that since the triangle is isosceles, the distance of the two points will be the same from y axis. If the coordinates of A are (x,y), the coordinates of B will be (-x,y). So, one coordinate will be (0,9) [vertex], (x,y) [A], and (-x,y) .

Now, we know that [tex]y=9-x^2[/tex] and the area of a triangle can be given by the determinant, [tex]D=\frac{1}{2}(0 9 1, x 9-x^2 1, -x 9-x^2, 1)[/tex]. (Sorry I don't know the tex for a determinant, hopefully someone who does could format it. Thank you.)

Solving this should be easy, specially if you use the properties of a det.
 
  • #4
But the real question was about the case where the "base" of the triangle, the non-congruent line- was NOT parallel to the x-axis.
 
  • #5
Sorry. My mistake. My case only works for the base parallel to the x axis.
 
  • #6
For the other case, if you assume the coordinates of B to be [tex](x_2,y_2)[/tex] and of A to be [tex](x_1,y_1)[/tex] and since the vertex is at the origin, assuming OB to be the base, OA comes out to be [tex]\sqrt{x_1^2+y_1^2}[/tex].

The equation of the line OA comes out to be [tex]y=(\frac{y_1}{x_1})x[/tex]. The distance of the point B from OA is [tex]\frac{y_2-\frac{y_1}{x_1}x_2}{\sqrt{1+\frac{y_1^2}{x_1^2}}}[/tex].

After solving for [tex]\frac{1}{2}\frac{y_2-\frac{y_1}{x_1}x_2}{\sqrt{1+\frac{y_1^2}{x_1^2}}}\sqrt{x_1^2+y_1^2}[/tex]

The area is given by [tex]\frac{1}{2}(x_1y2-y_1x_2)[/tex]. If you could find a relation between [tex]x_1, y_1[/tex] and [tex]x_2,y_2[/tex] you would be home free... how would you do that though?
 

What is an isosceles triangle in a parabola?

An isosceles triangle in a parabola is a triangle that has two sides of equal length and is found within the curve of a parabola. The base of the triangle is formed by a chord of the parabola, and the two equal sides are formed by tangents to the parabola at the endpoints of the chord.

How do you identify an isosceles triangle in a parabola?

To identify an isosceles triangle in a parabola, you can look for a chord of the parabola that is perpendicular to the axis of symmetry. The two tangents at the endpoints of the chord will intersect at the vertex of the parabola, forming an isosceles triangle.

What is the relationship between the vertex of a parabola and an isosceles triangle within it?

The vertex of a parabola is the point where the two tangents to the parabola intersect. This point is also the apex of the isosceles triangle formed by the tangents and the chord. Therefore, the vertex of the parabola is also the vertex of the isosceles triangle.

How do you find the area of an isosceles triangle in a parabola?

To find the area of an isosceles triangle in a parabola, you can use the formula A = 1/2 * b * h, where b is the length of the base (chord of the parabola) and h is the height of the triangle (the distance from the vertex to the base). You can also use the formula A = 1/2 * a * c, where a and c are the lengths of the equal sides of the triangle and c is the distance from the midpoint of the base to the vertex.

What is the significance of isosceles triangles in parabolas?

Isosceles triangles in parabolas have a few significant properties. They are always inscribed within the curve of the parabola, they have a vertex at the vertex of the parabola, and their area is always equal to 1/2 of the area of the rectangle formed by the base and the height of the parabola. These properties are useful in solving problems involving parabolas and can also be applied in other areas of mathematics.

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