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

Question

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.

Answer 1

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.

Answer 2

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.