Expert: Josh - 2/3/2004

Question
Hi, I know how to solve this problem but there is one thing I don't understand.
Write the equation of the ellipse with the given characteristics:
Vertices(2,-6) and (2,2) and minor axis of length 3.
I know how to solve the vertices but I don't understand the minor axis of length 3.
Do you count three horizontally from the center?  

Answer
Hi Nian,

The minor always refers to the distance between the extremities (vertices) of the ellipse on the shorter of the two sides. The question is whether the minor is parallel to the X-axis or the Y-axis.

To start with, you know that the standard ellipse equation is (X^2/a^2)+(Y^2/b^2)=1. If the ellipse is centered about (c,d), then, we have

(x-c)^2/(a^2)+(y-d)^2/(b^2)=1. ...[#1]

The distance between the two points P=(2,-6), Q=(2,2) on the axis of symmetry is given by |P-Q|=8. We cannot do anything to change this. According to the diagram given at the website (http://mathworld.wolfram.com/Ellipse.html), 2b=8 and b=4.

The midpoint of PQ (center of the ellipse) is therefore M=(c,d)=(2,-2)...[#2]

In this case, the major (maximum distance subtended by the ellipse) is found along the Y-axis, instead of the X-axis.
So, the minor is parallel to the X-axis. In particular, if we let R=(r,-2), S=(s,-2), the required distance between R and S, |R-S|=2a=3.

Using [#2] and knowing that "a=3/2", "b=4", the equation [#1] becomes (x-2)^2/(9/4)+(y+2)^2/(16)=1.

You can verify that point P and Q satisfy this equation by direct substitution.

Check that
(2-2)^2/(9/4)+(2+2)^2/(16)=1
and (2-2)^2/(9/4)+(-6+2)^2/(16)=1 yourself.

Cheers.