Question

need help with math

Answer 1

What is the equation of the ellipse with foci (0, 3), (0, -3) and co-vertices (1, 0), (-1, 0)?

Answer 2

Well, we can tell that the distance between a co-vertex and a focus is
$$\sqrt{(\pm3)^2 + (\pm1)^2} = \sqrt{9 + 1} = \sqrt{10}
$$Now, from the definition of an ellipse we know that the sum of distances from the foci has to be equal for all points. From our co-vertices we know that the sum has to be \(\sqrt{10} + \sqrt{10} = 2\sqrt{10}\)
Now let's take the vertices. The vertices lay on the same line as the foci, and from the symmetry of the ellipse it's not hard to see that the sum of distances of each vertex from the foci is equal to the distance between the two vertices.
Which means that the distance of each vertex from from the origin is half that distance and therefore is \(\sqrt{10}\).
So we know that our vertices are at \((\pm\sqrt{10},0)\). So by plugging in:
$$
x^2 + (ky)^2 = r^2 \\
x^2 + k^2 \cdot y^2 = r^2 \\
(\pm\sqrt{10})^2 + k^2 \cdot (0)^2 = r^2 \\
r^2 = 10 \\
$$And by plugging our co-vertices at \((0, \pm1)\) we get:
$$
(0)^2 + k^2 \cdot (\pm 1)^2 = 10 \\
k^2 = 10
$$ So our formula is:
$$
x^2 + 10 \cdot y^2 = 10
$$
http://www.wolframalpha.com/input/?i=x%5E2+%2B+10+y%5E2+%3D+10
You can also look at the 'properties' of the 'Geometric figure' there and see the properties of the ecllipse.