Allright. First off, draw a diagram and use parametric form because these questions are generally easier with it. I'll type up a worked solution- give me a few minutes.
<a href="http://www.codecogs.com/eqnedit.php?latex=\textup{Note the three points }P(acos\theta,bsin\theta), S(ae,0), M(\frac{a(cos\theta@plus;e)}{2},\frac{bsin\theta}{2})\\ \textup{We want to eliminate the theta's from the M value}\\ x=\frac{a(cos\theta@plus;e)}{2}, y=\frac{bsin\theta}{2}\\ \textup{Remember that }sin^2\theta@plus;cos^2\theta=1 \textup{ and rearranging the above to use this: }\\ cos\theta=\frac{2x-ae}{a}, sin\theta=\frac{2y}{b}\\ \frac{(2x-ae)^2)}{a^2}@plus;\frac{4y^2}{b^2}=sin^2\theta@plus;cos^2\theta=1\\ \textup{which is an ellipse centre }\frac{ae}{2},0" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\textup{Note the three points }P(acos\theta,bsin\theta), S(ae,0), M(\frac{a(cos\theta+e)}{2},\frac{bsin\theta}{2})\\ \textup{We want to eliminate the theta's from the M value}\\ x=\frac{a(cos\theta+e)}{2}, y=\frac{bsin\theta}{2}\\ \textup{Remember that }sin^2\theta+cos^2\theta=1 \textup{ and rearranging the above to use this: }\\ cos\theta=\frac{2x-ae}{a}, sin\theta=\frac{2y}{b}\\ \frac{(2x-ae)^2)}{a^2}+\frac{4y^2}{b^2}=sin^2\theta+cos^2\theta=1\\ \textup{which is an ellipse centre }\frac{ae}{2},0" title="\textup{Note the three points }P(acos\theta,bsin\theta), S(ae,0), M(\frac{a(cos\theta+e)}{2},\frac{bsin\theta}{2})\\ \textup{We want to eliminate the theta's from the M value}\\ x=\frac{a(cos\theta+e)}{2}, y=\frac{bsin\theta}{2}\\ \textup{Remember that }sin^2\theta+cos^2\theta=1 \textup{ and rearranging the above to use this: }\\ cos\theta=\frac{2x-ae}{a}, sin\theta=\frac{2y}{b}\\ \frac{(2x-ae)^2)}{a^2}+\frac{4y^2}{b^2}=sin^2\theta+cos^2\theta=1\\ \textup{which is an ellipse centre }\frac{ae}{2},0" /></a>