Answer:
Step-by-step explanation:
This is really hard to try and explain over the internet here, but I'll do my best, assuming you have some experience with ellipses.
We are given the center of (3, 2), and the focus of (6, 2). The focus is 3 units to the right of the center; the other focus is 3 units to the left of the center at (0, 2). The focus has coordinates of (-ae, 2) and (ae, 2) from left to right and the distance between them then is 2ae. There are 6 units between the 2 foci, so 2ae = 6.
The directrix is 8 1/3 units from the center to the right AND the left. The directrices have distances of 2a/e between them. There are 16 2/3 units between the 2 directrices, so
[tex]\frac{2a}{e}=\frac{50}{3}[/tex]
We can solve for the eccentricity of this ellipse given the fact that the eccentricity is the ratio of the distance between the foci to the distance between the directrices. Therefore,
[tex]\frac{2ae}{2\frac{a}{e} }=\frac{6}{\frac{50}{3} }[/tex]
The 2's andd the a's cancel, leaving you with
[tex]e^2=\frac{18}{50}=\frac{9}{25}[/tex]
That gives us then that
[tex]e=\frac{3}{5}[/tex]
Going back to the identity for the distance between the foci, 2ae = 6, we fill in e and solve for a:
[tex]2a(\frac{3}{5})=6[/tex] and
[tex]\frac{6a}{5}=6[/tex]
which gives us that a = 5
Now use the identity [tex]b^2=a^2(1-e^2)[/tex] to solve for b:
[tex]b^2=25(1-\frac{9}{25})[/tex] and
[tex]b^2=25(\frac{16}{25})[/tex] and
b = 4
Because this is a horizontally stretched ellipse, it is of the form
[tex]\frac{(x-h)^2}{a^2} +\frac{(y-k)^2}{b^2} =1[/tex]
we fill in as
[tex]\frac{(x-3)^2}{25}+\frac{(y-2)^2}{16}=1[/tex]
Again, since you are expected to be able to solve for the equation of an ellipse at this advanced level, I am assuming that the equations I gave above as a means to solve for the different characteristics of an ellipse are familiar to you. This is definitely NOT beginning conics!
