Distance from focus to nearest point in ellipse

Question

Consider an ellipse with semi-axes $a$ (major) and $b$ (minor). For such an ellipse the distance of focus to the centre is:

$f = \sqrt{a^2-b^2}$

Now, the distance from the focus to the nearest point on the ellipse is along the major semi-axis a, thus this distance is:

$r_1 = a - f = a - \sqrt{a^2-b^2}$

Two simple questions now:

• How can we prove this is the shortest distance?

• Can we somewhat prove that the following is always true:

$\frac{a - \sqrt{a^2-b^2}}{b} < 1$

Answer 1

If $a/b=t>1,$

$$t-\sqrt{t^2-1}=\dfrac1{t+\sqrt{t^2-1}}<\dfrac1t<1$$

Alternatively, $t=\csc2y,0<2y\le\dfrac\pi2$

Answer 2

For 1) I wanted to do a very simple thing. A top part of ellipse has the equation

$y = b\sqrt{1-\frac{x^2}{a^2}}$

The squared distance from focus to any point is then given by

$d^2 = \left(x - \sqrt{a^2-b^2}\right)^2 + b^2\left(1-\frac{x^2}{a^2}\right)$.

If now we compute the derivative:

$\frac{\mathrm{d}}{\mathrm{d}x} \left(d^2\right) = 2 \left[ \left(x - \sqrt{a^2-b^2}\right) - \left(\frac{b}{a}\right)^2 x \right] = 2 \left[ x \left( 1 - \frac{b^2}{a^2} \right) - \sqrt{a^2-b^2} \right] = 0$

from that we should get:

$x = \frac{\sqrt{a^2-b^2}}{1 - \frac{b^2}{a^2}}$

But there must be some mistake here since for minimum $x = a$ and for maximum $x = -a$. Many thanks in advance.