Ellipse question....?

Question

I am totally lost on this problem. Could somebody please assist me? Thank you very much!

The planets move around the sun in elliptical orbits with the sun at one focus. The point in the orbit at which the planet is closest to the sun is called perihelion, and the point at which it is 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. Find an equation for the earth's orbit. (Place the origin at the center of the orbit with the sun on the x-axis.)

Here's all I have so far:

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

so...

(x^2 / 2.25 x 10^16) + y^2 / ???) = 1

Basically, I need help on the ??? part. Thanks again!

Answer 1

ok so you have a the half big axis
b the half small axis and c the half focal distance.
a+c=153... and a-c= 147 so a= 150 and a^2 = 225... Now
b^2=a^2-c^2=(a-c)(a+c)= 22500 - 9=2.2491 10^16. This is your ??? Ok?

Answer 2

Make a sketch, ellipse, center at origin, focus a little to the right on the x axis. Label focus with a c, right x intercept with a, top y intercept with b. Perihelion distance is a-c; aphelion distance is a+c.

So a = (153 million + 147 million)/2 = 150 million. c = (153 million - 147 million)/2 = 3 million. and b²+c² = a², so b² = 150 million ² - 3 million ² = 2.2491 x 10^16, which goes under y². Eccentricity is c/a = 3/150 = 1/50 = 0.02, which is close to zero and tells us the orbit is close to circular. If that were not true, perihelion and aphelion temperatures would be very different, and life might well not exist.

Answer 3

The sun has coordinates (c,0)
When the earth is at perihelion its coordinates would be (a-c ,0) where a-c=147*10^6 km
At aphelion a+c= 153*10^6 km
Sum up both and you get 2a= 300*10^6km
so a = 150*10^6 km
substract and 2c= 6*10^6km
so
c= 3*10^6 km
as b^2 = a^2-c^2 = 10^12*(22,491) The equation would be

x^2/(22,500*10^12) +y^2/(22491*10^12)=1

Very close to a circunsference

Answer 4

You know the focii, the other focus will be 147000000 from aphelion. Since you know the focus, find the center, and more importantly the distance from the center. Then use that to find b. I know there is an equation linking distance from center to focus, a, and b.

Answer 5

a = 1/2 (153,000,000 + 147,000,000) = 150, 000, 000
f = 1/2 (153,000,000 - 147,000,000) = 3, 000, 000


No, you need original definition of ellipse: the geometric
location of points E for which
|F1 E| + |F2 E| = D = const

Thus basiacally you place
F1 at (f, 0),
F2 at (-f, 0),
and write the equation

sqrt((x-f)^2 + y^2) + sqrt((x+f)^2 + y^2) = 2a
sqrt((x-f)^2 + y^2) = 2a - sqrt((x+f)^2 + y^2)
[sqrt((x-f)^2 + y^2)]^2 = [2a - sqrt((x+f)^2 + y^2)]^2
(x-f)^2 + y^2 = 4a^2 - 4a sqrt((x+f)^2 + y^2) + (x+f)^2 + y^2

4a^2 - 4a sqrt((x+f)^2 + y^2) + 4fx = 0
a + f/a x = sqrt((x+f)^2 + y^2)
a + f/a x = sqrt((x+f)^2 + y^2)

a^2 + 2afx + (f/a)^2 x^2 = (x+f)^2 + y^2
a^2 + 2fx + (f/a)^2 x^2 = x^2 +2fx + f^2 + y^2
a^2(a + f)(a -f) = (a + f)(a-f) x^2 + y^2
1 = x^2/a^2 + y^2/ [(a + f)(a-f)]

thus b = sqrt[(a + f)(a-f)]