Elipse word problem, Very confusing

Question

The distance of Amateru from its star, epsilon Tauri, ranges from  million mi to  million mi. Find the equation of its elliptical orbit where the star is one of the foci. Assume that the center is the origin and its major axis is the y axis

Answer 1

You are missing the numbers in your question, which can happen sometimes when you copy a question directly into Wyzant. I'll call the larger of the two numbers u and the smaller of the two numbers v. You'll have to substitute those values into the final answer I give.

It will help a great deal to have a sketch handy. You'll also need a few properties of an ellipse. Namely, the distance between the center of an ellipse and its longest end is the semi-major axis A, the distance between the center of an ellipse and either of its foci is C, and C is related to A and B (the semi-minor axis) by C²=A²-B². Lastly, the general equation for an ellipse where the semi-major axis is on the y-axis is

x²/B²+y²/A²=1

So our task boils down to finding A and B.

When the distance between Amateru and epsilon Tauri is at its greatest, the planet is at the point on the ellipse that is the farthest it can possibly be from one of its foci. If you made a good sketch and labelled A and C properly, you should be able to see that

A+C=u.

On the other hand, when the distance between Amateru and epsilon Tauri is smallest, the planet is at the point on the ellipse that is the closest it can possibly be to one of its foci. You should then find that

A-C=v.

This is a system of equations that we can solve for A and C. When they're as simple as this, I prefer to add and subtract the equations to cancel one of the two variables and then solve. You should find

A=1/2 (u+v)

C=1/2 (u-v).

We've got A, so now we need to use C to get B. From C²=A²-B², we can solve for B:

B=sqrt(A²-C²).

To avoid rounding errors, I'll get this in terms of u and v by substituting what I just found:

B=sqrt(A²-C²)

=sqrt([1/2(u+v)]²-[1/2(u-v)]²)

=sqrt(1/4[u²+2uv+v²+u²-2uv+v^2])

=sqrt([u²+v²]/2)

So, now we have

A=(u+v)/2

B=sqrt([u²+v²]/2)

You can substitute the missing u and v values to get these, and then you have the equation for your ellipse

Answer 2

Since you haven't given the data, I'll just answer generally.

By giving the two extreme points, r_a and r_p (a for aphelion and p for perihelion), you can derive a, b, and c for the ellipse. The semimajor axis (a) is from the center to these points and the focal length (c) is the distance from the center to one of the foci. From the geometry, you know that r_a = a+c and r_p = a - c . We can solve for a and c: a = (r_a + r_p)/2 and c = (r_a - r_p)/2 and you can solve for b using the ellipse relation a² = b²+c².