1

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:

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

rk85
  • 127

2 Answers2

2

If $a/b=t>1,$

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

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

lab bhattacharjee
  • 274,582
0

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.

rk85
  • 127
  • 1
    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