# Geometric proof that the vertex of an ellipse is the closest point to its closest focus

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.

• 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: $r=\frac{p}{1+e \cos \theta}$ Apr 22 at 12:19
• 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 $P$. Apr 22 at 12:36

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.
• 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 $(a,0)$... Apr 22 at 17:03
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.