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Question Number 130358 by ajfour last updated on 24/Jan/21

Commented by ajfour last updated on 24/Jan/21

$T w o i d e n t i c a l e l l i p s e s .$ $F i n d m a x i m u m v a l u e o f b / a .$
Commented by MJS_new last updated on 25/Jan/21

$once again thinking backwards (exchanging$ $the ellipses)$ $let a = 1$ $(1) search 2 tangents to the 1^{st} ellipse with$ $t_{1} ⊥ t_{2}$ $(2) shift and rotate the ellipse to get the 2^{nd}$ $ellipse with center at t_{1} \cap t_{2} and axes t_{1}$ $and t_{2}$ $[the intersections of all possible rectangular$ $pairs of tangents form a circle with radius$ $\sqrt{b^{2} + 1} and center (\begin{matrix} 0 \\ 0 \end{matrix})]$ $I think we get the maximum b when t_{1} and$ $t_{2} are symmetric to x = 0 \Leftrightarrow t_{1} \cap t_{2} = (\begin{matrix} 0 \\ \sqrt{b^{2} + 1} \end{matrix})$ $in this case the equation of the 2^{nd} ellipse is$ $easy to get because the angle of rotation is$ $\frac{π}{4} . now we must intersect the ellipses and$ $find b in order to get exactly one intersection .$ $this leads to a 4^{th} degree polynome for x .$ $I once posted the conditions for all possible$ $kinds of solutions for this . it leads to a 12^{th}$ $degree polynome for b^{2} which has exactly$ $one real solution we can only approximate .$ $I get b \approx . 368 671 126 255$ $not sure if my idea is right . . . but it looks a$ $lot to me we will get a smaller value for b when$ $using a different pair of tangents . . .$
Commented by MJS_new last updated on 25/Jan/21

$the equations for plotting are :$ $ellipse 1 : y = \pm b \sqrt{1 - x^{2}}$ $tangents : y = \pm x + \sqrt{b^{2} + 1}$ $ellipse 2 : y = \frac{1 - b^{2}}{b^{2} + 1} x + \sqrt{b^{2} + 1} \pm \frac{b \sqrt{2 b^{2} + 2 - 4 x^{2}}}{b^{2} + 1}$ $it seems obvious that$ $(1) if b gets smaller the ellipses will lose contact$ $\Rightarrow we have to rotate the tangents$ $(2) if b gets larger we get 2 intersections$ $\Rightarrow we have to rotate the tangents but then$ $the space gets even smaller$ $[this needs a proof . . .]$
Commented by MJS_new last updated on 25/Jan/21

$above (1) obviously we rotate to the left$ $\Rightarrow (2) we cannot rotate to the left with$ $larger b but rotating to the right seems no$ $good idea either [plot it!]$
Commented by mr W last updated on 25/Jan/21

$s a y μ = \frac{b}{a}$ $μ_{m a x} c a n n o t b e o b t a i n e d e x a c t l y .$ $i n a n e a r l i e r q u e s t i o n Q 127509 i g o t$ $μ_{m a x} \approx 0.38436919447$ $w h i c h o c c u r s n o t a t a n a n g l e \frac{π}{4} .$
Commented by mr W last updated on 25/Jan/21

Commented by MJS_new last updated on 25/Jan/21

$I see .$ $does the y - axis of the red ellipse intersect$ $the green one at (\begin{matrix} 0 \\ - b \end{matrix}) ?$
Commented by mr W last updated on 25/Jan/21

$n o . w e c a n s e e t h i s i n t h e d i a g r a m .$
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