Question

What is the ellipse which has vertices at `v_1 = (5,10)` and `v_2=(-2,-10)`, passing by point `p_1=(-5,-4)`?

Answer 1

Please see the explanation.

Explanation:

To go from `v_1` to `v_2` the x coordinate decreased by 7 and the y coordinate decreased by 20.

To go from `v_1` to the center `(h, k)`, the x coordinate must decrease by 3.5 and the y coordinate must decrease by 10:

Therefore, `h = 1.5, k = 0,` and the center of the ellipse is `(1.5, 0)`

The length of the semi-major axis, a, is the distance from the center to either vertex:

`a = sqrt((5 - 1.5)^2 + (10 - 0)^2)`

`a = sqrt((3.5)^2 + 10^2)`

`a = sqrt(449)/2`

Let A = the angle of rotation, then:

`sin(A) = 10/(sqrt(449)/2), and cos(A) = 3.5/(sqrt(449)/2)`

Rationalizing both denominators:

`sin(A) = (20sqrt(449))/449 and cos(A) = (7sqrt(449))/449`

Here is a reference for An Rotated Ellipse that I not at the origin

`((x - h)cos(A) + (y - k)sin(A))^2/a^2 + ((x - h)sin(A) - (y - k)cos(A))^2/b^2 = 1`

Solving for `b`

`1 - ((x - h)cos(A) + (y - k)sin(A))^2/a^2= ((x - h)sin(A) - (y - k)cos(A))^2/b^2`

`b^2 = (a^2((x - h)sin(A) - (y - k)cos(A))^2)/(a^2 - ((x - h)cos(A) + (y - k)sin(A))^2)`

`b = sqrt((a^2((x - h)sin(A) - (y - k)cos(A))^2)/(a^2 - ((x - h)cos(A) + (y - k)sin(A))^2))`

Force this to contain the point `(-5, -4)`:

`b = sqrt(((449/4)((-5 - 1.5)(20sqrt(449))/449 - (-4 - 0)(7sqrt(449))/449)^2)/((449/4) - ((-5 - 1.5)(7sqrt(449))/449 + (-4 - 0)(20sqrt(449))/449)^2))`

I used WolframAlpha to do the evaluation:

`b = 5.80553`

Here is the final equation:

`((x - 1.5)(20sqrt(449))/449 + (y - 0)(7sqrt(449))/449)^2/(sqrt(449)/2)^2 + ((x - 1.5)(7sqrt(449))/449 - (y - 0)(20sqrt(449))/449)^2/(5.80553)^2 = 1`

Here is a graph to prove it: