# Thread: Find centre of ellipse (Polar)

1. ## Find centre of ellipse (Polar)

Show that the centre C of the ellipse $\displaystyle r = \frac{ep}{1 + ecos(\theta)}$ has the coordinates $\displaystyle \lgroup \frac{e^2p}{1 - e^2}, -\pi \rgroup$

Where e is the eccentricity and p the distance from focus to directrix.

Here is my attempt

The vertices of the major axis occur at 0 and $\displaystyle \pi$

The vertices of the minor axis occur at $\displaystyle \frac{\pi}{2}$ and $\displaystyle \frac{-\pi}{2}$

First coordinate by midpoint rule $\displaystyle \frac{\frac{ep}{1 - e} + \frac{ep}{1 + e}}{2} = \frac{\frac{2ep}{1 - e^2}}{2} = \frac{ep}{1 - e^2}$

By similar reasoning, second coordinate by midpoint rule yields ep.

So I get $\displaystyle C\lgroup \frac{ep}{1 - e^2}, ep \rgroup$

Where have I gone wrong?

2. ## Re: Find centre of ellipse (Polar)

Originally Posted by Altmer
Show that the centre C of the ellipse $\displaystyle r = \frac{ep}{1 + ecos(\theta)}$ has the coordinates $\displaystyle \lgroup \frac{e^2p}{1 - e^2}, -\pi \rgroup$

Where e is the eccentricity and p the distance from focus to directrix.

Here is my attempt

The vertices of the major axis occur at 0 and $\displaystyle \pi$

The vertices of the minor axis occur at $\displaystyle \frac{\pi}{2}$ and $\displaystyle \frac{-\pi}{2}$

First coordinate by midpoint rule $\displaystyle \frac{\frac{ep}{1 - e} + \frac{ep}{1 + e}}{2} = \frac{\frac{2ep}{1 - e^2}}{2} = \frac{ep}{1 - e^2}$

By similar reasoning, second coordinate by midpoint rule yields ep.

So I get $\displaystyle C\lgroup \frac{ep}{1 - e^2}, ep \rgroup$

Where have I gone wrong?
You are attempting to use the midpoint formula (for x-y coordinates) for polar coordinates

3. ## Re: Find centre of ellipse (Polar)

Originally Posted by Altmer
Show that the centre C of the ellipse $\displaystyle r = \frac{ep}{1 + ecos(\theta)}$ has the coordinates $\displaystyle \lgroup \frac{e^2p}{1 - e^2}, -\pi \rgroup$

Where e is the eccentricity and p the distance from focus to directrix.

Here is my attempt

The vertices of the major axis occur at 0 and $\displaystyle \pi$

The vertices of the minor axis occur at $\displaystyle \frac{\pi}{2}$ and $\displaystyle \frac{-\pi}{2}$

First coordinate by midpoint rule $\displaystyle \frac{\frac{ep}{1 - e} + \frac{ep}{1 + e}}{2} = \frac{\frac{2ep}{1 - e^2}}{2} = \frac{ep}{1 - e^2}$

By similar reasoning, second coordinate by midpoint rule yields ep.

So I get $\displaystyle C\lgroup \frac{ep}{1 - e^2}, ep \rgroup$

Where have I gone wrong?
subtract rather than add since the second vertex is on the negative x-axis