Htykuut

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, $t=csc2y,0<2yledfracpi2$

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

$y = bsqrt{1-frac{x^2}{a^2}}$

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

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

If now we compute the derivative:

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

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.