The question is as follows:

Reflection property of the ellipse. Suppose that $P$ is a point on an ellipse whose focal points are $F_1$ and $F_2$. Draw the intersecting lines $PF_1$ and $PF_2$, as well as the bisectors of the four angles they form. Consider the bisector that does not separate $F_1$ and $F_2$. Prove that given any point $Q$ other than $P$ on this line, $QF_1 + QF_2 > PF_1 + PF_2$. Explain why the line meets the ellipse only at $P$. Justify the title of this problem.

I drew the angle bisectors of the four angles that are formed after intersecting the lines extending from $PF_1$ and $PF_2$. It looks like from the drawing that the bisector that is not separating the foci is the line tangent to the ellipse at $P$. I, however, can't prove that that statement is true. Any help will be greatly appreciated.

  • 2
    $\begingroup$ It would help if you explain what definition of ellipse you are using, because there are several different ways to answer this question, depending on the definition. Are you defining ellipses using analytic geometry? Are you defining them in the classical manner where $|PF_1|+|PF_2|$ is required to be a constant? $\endgroup$
    – Lee Mosher
    May 28 '18 at 19:27


If $F_2'$ is the reflection of $F_2$ about the line, then it is easy to show that $PF_1+PF_2=F_1F_2'$ and $QF_1+QF_2>F_1F_2'$.


A point $P$ lies on the ellipse if $PF_1+PF_2=2a$. A point $Q$ such that $QF_1+QF_2>2a$ lies outside the ellipse.


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