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It is well known that, in an ellipse $$ \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1, \qquad a>b,$$ the vertex $A(a, 0)$ is the closest point in the ellipse to the focus $F(c, 0)$ and the farthest from the other focus $F'(-c, 0).$

It's easy to prove this by differentiating the distance function, but I've never seen a geometric proof of this.

Thanks.

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  • $\begingroup$ There is another very simple analytical mean to prove your result by using the polar equation of an ellipse when the origin is at one of its foci: $r=\frac{p}{1+e \cos \theta}$ $\endgroup$
    – Jean Marie
    Apr 22 at 12:19
  • $\begingroup$ Let $l$ is distance from some point $P$ of ellipse from $F$, then distance from $F'$ is $2a-l$. Using triangle rule in triangle $FF'P$: $2a-l \geq 2c+l \Rightarrow l \geq a-c$ with equality only if $F$ is between $F'$ and $P$. $\endgroup$ Apr 22 at 12:36

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COMMENT.- Despite "closest" and "farthest" are rather analytical notions (minimum, maximum), it seems you can solve geometrically your question. Take $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2} = 1, a>b$ and a circle with center $(\sqrt{a^2-b^2},0)$ and radius $a-\sqrt{a^2-b^2}$.The adequated reasoning becomes.enter image description here

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  • $\begingroup$ After "the adequate reasoning becomes" there is some void... It should be said (because the picture is not a proof) that the circle with equation $x^2+y^2-2 \sqrt{a^2-b^2}x=a^2-2a \sqrt{a^2-b^2}$ as no other common point with the ellipse than point $(a,0)$... $\endgroup$
    – Jean Marie
    Apr 22 at 17:03
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    $\begingroup$ @Jean Marie: I haven't said anything about using the definitions of focus and eccentricity of the ellipse in my comment so that the OP can realize and reason about his problem. The attached figure (which is why I didn't put this in the comments section!) is intended to suggest to the OP what to do and is not intended to prove anything at all (a line passing through the center of what? cuts both curves at distinct points except in one point, etc). $\endgroup$
    – Piquito
    Apr 24 at 15:08
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Let $x,y$ be the respective distances from the focii.By definition $$x+y=2a$$

As $2c$ is the third side of triangle $PF_1F_2$, By triangle inequality, $$2c\geq x-y$$ Substitute from above, $$x-y=2a-2y$$ Putting in above inequality, $$y\geq a-c$$ By equality , we get $y=a-c$ which is the point that you need. Note that I assumed $x>y$ so a minimum y means that $y$ is closer to the required focus, simultaneously $x$ has achieved a maximum. This result is a consequence of constant sum of distances from 2 points and the triangle inequality,So you would not need to use differentiation.

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  • $\begingroup$ (+1) Excellent answer ! At the same time Ivan Kaznacheyeu has had the same idea ! $\endgroup$
    – Jean Marie
    Apr 22 at 21:33
  • $\begingroup$ @sebpar You should consider this answer which uses only the focal definition of an ellipse. $\endgroup$
    – Jean Marie
    Apr 22 at 21:35

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