Ellipse through three points

[Nelourir]

Homework Statement

First of all, I'm not sure the following is achievable at my level (last year of high school), as it is not posed as a direct question, but if yes, could you point me in the right direction?

I'm trying to find an ellipse that goes through three points, (0,0) , (b,0) and (b/2,h).
This actually represents a building with a base of width b and a height of h.

The building is symmetrical, therefore the ellipse is not 'rotated' (I'm not sure how to express this mathematically).

The Attempt at a Solution

Using the equation for an ellipse (not trying to confuse anyone but h is not H and b is not B):
(((X-H)^2)/(A^2)) + (((Y-K)^2)/(B^2)),
I find the following (using logic and no algebra):

H is b/2 as the centre of the ellipse always goes through the line x=b/2
A is greater than or equal to b/2. (otherwise the ellipse cannot touch both points on the x axis)
B is greater than or equal to h. (I don't want the building to have angles greater than 90^o)
Using the above statement, I expressed B as h + C and found that K = -C.

I realize there is an infinite number of ellipses (at least I'm assuming so), but I think that A should be dependent on b, h and/or C. Could you help me find this?

Thank you for your help.

EDIT: OK, I think I found my answer by replacing A by b/2 + D, plugged in x=0, y=0 then I rearranged until it looked like a quadratic equation with D as the 'variable' and I got an answer for D in terms of b, h and D. Sorry for making a new topic :shy:.

 

[mighty2000]

Thank you for your post. Your question is definitely achievable at your level, and it is great that you are trying to find a solution using logic and algebra.

To find an ellipse that goes through three points, you can use the general form of an ellipse equation ((x-h)^2/a^2 + (y-k)^2/b^2 = 1) and substitute in the coordinates of the three points. This will give you three equations with three unknowns (a, b, and k). You can then solve for these unknowns using algebraic manipulation and substitution.

In your attempt at a solution, you have correctly identified that the center of the ellipse must go through the line x=b/2, and that b/2 will be the value of h in the equation. However, A and B do not have to be greater than or equal to b/2 and h respectively. In fact, A and B can be any positive values as long as they satisfy the equation.

To find a unique ellipse, you will need to use additional information or constraints. For example, you mentioned that the building is symmetrical, so you can use that to determine the value of A and B. Additionally, you can use the fact that the ellipse must have a maximum height of h, so this can help you determine the values of a and b.

I would suggest revisiting your solution and using the equations of the three points to solve for A and B. You can also try using the symmetry of the building to further constrain your solution. I hope this helps. Good luck with your problem!