Last question on ellipses? You got tired of them?

Imagine, or plot those two given points on an x,y set of coordinates. So one side of the triangle is 10 units long. The other two sides should be 30-10 = 20units combined.

Now we play. Place the 3rd corner along the y-axis. No triangles here. But the 3rd "corner" is 5 units above "corner" (0,5) or 5 units below "corner" (0,-5). So the vertical axis, 2b, of the ellipse is 20 units long.

Place the 3rd corner on the, say, positive x-axis. We have an isosceles triangle, made up of two congruent right triangles.

In any of the right triangles: hypotenuse = 20/2 = 10 units. Vertical leg = 5 units. So horizontal leg = sqrt(10^2 -5^2) = sqrt(75) units.

Meaning, the horizontal axis, 2a, of the ellipse is 2sqrt(75) units long.

The ellipse:

(x^2)/(a^2) +(y^2)/(b^2) = 1

(x^2)/(sqrt(75))^2 +(y^2)/(10^2) = 1

(x^2)/75 +(y^2)/100 = 1 ------------------------answer.