## write eqn of ellipse that models path of Halley's comet

Complex numbers, rational functions, logarithms, sequences and series, matrix operations, etc.
Stranger_1973
Posts: 21
Joined: Sun Feb 22, 2009 9:56 pm

### write eqn of ellipse that models path of Halley's comet

Halley's Comet orbits the sun every 76 years. The comet travels in an elliptical path, with the sun at one of the foci. At the closest point, or perihelion, the distance of the comet to the sun is 8.8 x 10^7 km. At the furthest point, or aphelion, the distance of the comet from the sun is 5.3 x 10^9 km. Write an equation of the ellipse that models the path of Halley's Comet. Assume the sun is on the x-axis.

They don't give me the radius of the sun, so should I treat it as just a point?

I put the center at the origin. The closest distance will be "a" less "c". The other distance will be "c" plus "a". So I have:

8.8 x 10^7 = a - c
5.3 x 10^9 = a + c

Adding gives me 8.8 x 10^7 + 5.3 x 10^9 = 2a, so (8.8 + 530) x 10^7 = 2a, a = 269.4 x 10^7 = 2.694 x 10^9. Then c = 2.606 x 10^9. a^2 - c^2 = b^2, so b = 0.4664 x 10^9.

Am I doing this right?

Stranger_1973
Posts: 21
Joined: Sun Feb 22, 2009 9:56 pm

### Re: write eqn of ellipse that models path of Halley's comet

bump

stapel_eliz
Posts: 1737
Joined: Mon Dec 08, 2008 4:22 pm
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Stranger_1973 wrote:Halley's Comet orbits the sun every 76 years. The comet travels in an elliptical path, with the sun at one of the foci. At the closest point, or perihelion, the distance of the comet to the sun is 8.8 x 10^7 km. At the furthest point, or aphelion, the distance of the comet from the sun is 5.3 x 10^9 km. Write an equation of the ellipse that models the path of Halley's Comet. Assume the sun is on the x-axis.

They don't give me the radius of the sun, so should I treat it as just a point?

I would guess so, but you might want to say that on your homework: "In the following, the sun is regarded as a point, since no radius or diameter information is specified." That should "cover" you.

The numbers are nasty. You might find it simpler to disregard the "ten to whatever power" bit until the end. (What you've done is fine, but it's more prone to errors.) Your relationships are correct:

. . . . .perihelion: 8.8 = a - c
. . . . .aphelion: 530 = a + c
. . . . .(a - c) + (a + c) = 538.8 = 2a
. . . . .a = 269.4
. . . . .c = 260.6

Then:

. . . . .$a^2\, -\, c^2\, =\, b^2$

. . . . .$269.4^2\, -\, 260.6^2\, =\, 4664\, =\, b^2$

. . . . .$b\, =\, \sqrt{4664}\, \approx 68.29$

You don't actually need the value of "b", though, since you only need "b2" for the equation. So plug the values you've got into the basic equation for an ellipse:

. . . . .$\frac{x^2}{a^2}\, +\, \frac{y^2}{b^2}\, =\, 1$

(Remember to put the powers back in, though!)

Stranger_1973
Posts: 21
Joined: Sun Feb 22, 2009 9:56 pm

Thank you.