# Distance from focus to nearest point in ellipse

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:

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

If $$a/b=t>1,$$

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

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

• For $1)$ $$d^2=(ae-a\cos u)^2+(b\sin u-0)^2,b^2=a^2(1-e^2)$$ – lab bhattacharjee Nov 23 '18 at 15:24

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.

• You correctly have $\frac{d}{dx}(d^2) = 2\left(1 - \frac{b^2}{a^2}\right)(x - c)$, where $c = \frac{a^2}{\sqrt{a^2 - b^2}}$. But note that $c > a$, so the derivative of $d^2$ with respect to $x$ is strictly negative for all the possible values of $x$, i.e. all those between $-a$ and $a$. This actually confirms the conclusion that the minimum value of $d^2$ is attained when $x = a$. [Sorry if you tried to read this a couple of minutes ago - my thumb accidentally hit the Enter key!] – Calum Gilhooley Nov 23 '18 at 20:06
• Yes, that is correct. Thanks. – rk85 Nov 25 '18 at 10:30