Need help with ellipse problems! (math)?

Question

The planets move around the sun in elliptical orbits with the sun at one focus. The point in the orbit as which the planet is closest to the sun is called perihelion, and the point at which it is the farthest is called aphelion. These points are the vertices of the orbit. The earth's distance from the sun is 147,000,000 km at perihelion and 153,000,000 km at aphelion.

Now here's the question:

With an eccentricity of 0.25, Pluto's orbit is the most eccentric in the solar system. The length of the minor axis of its orbit is approximately 10,000,000,000 km. Find the distance between Pluto and the sun at perihelion and at apheliion.

Now here is the diagram:

http://img126.imageshack.us/img126/544/picnx2.jpg

Answer 1

Let a be half of the length of the major axis and b half of the length of the minor axis.
The eccentricity is defined as:
ε = √(a² - b²) / a = √(1 - (b/a)²)
ε and b are known. Hence:
a = b / √(1 - ε²) = 5,000,000,000km / √(1 - 0.25²) = 5,163,977,795km

The distance between the center of the ellipse and the foci is given by:
c = √(a² - b²) = ε · a

At the perihelion the distance from the vertex to the focus is
dp = a - c = (1 - ε) · a = 3,872,983,346km

At the aphelion the distance from the vertex to the focus is
dp = a + c = (1 + ε) · a = 6,454,972,244km

Answer 2

With an eccentricity of 0.25, Pluto's orbit is the most eccentric in the solar system. The length of the minor axis of its orbit is approximately 10,000,000,000 km. Find the distance between Pluto and the sun at perihelion and at aphelion.

Let
ε = eccentricity

The minor axis is 2b.

b = 10,000,000,000/2 km = 5,000,000,000 km

c = √(a² - b²)

ε = c/a = √(a² - b²) / a = √[(a² - b²)/a²]

ε² = (a² - b²)/a²
ε²a² = a² - b²
ε²a² - a² = -b²
a²(1 - ε²) = b²
a² = b²/(1 - ε²) = b²/(1 - .25²) = b²/(1 - 1/16) = 16b²/15

a² = 16(5,000,000,000)²/15
a = 20,000,000,000/√15 = 5,163,977,795

ε = c/a
c = aε = a/4

The distance between Pluto and the sun at perihelion is:

a - c = a - a/4 = (3/4)a = 3,872,983,346 km

The distance between Pluto and the sun at aphelion is:

a + c = a + a/4 = (5/4)a = 6,454,972,244 km