Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

so I have this problem for my homework:

Consider the elipse: $\dfrac {x^2}{a^2} + \dfrac{y^2}{b^2}=1$ where $0<b<a$. For every point (x,y) on the ellipse find the the perpendicular line to the ellipse so that the point $(x,y)$ is on that line. This line cuts the ellipse in another point $(x',y')$. Prove that the distance between these two points is $$D(x,y)=2\dfrac {(\dfrac {x^2}{a^4} + \dfrac{y^2}{b^4})^{3/2}}{\dfrac {x^2}{a^6} + \dfrac{y^2}{b^6}}\ .$$ So I've alrredy done that; but after that it says:

Use Lagrange Multipliers to minimize the function $D(x,y)$. But the equations get complicated and messy and I dont know if i have to consider the line or just the ellipse. Logically the minimum is at $(0,b)$.

If anyone can help me I would thak you so much!

share|improve this question
but 0 isn't in the elipse, the profersor wants ou to find the pont (x,y) on the elipse such that the distance to (x',y') is minumun –  Buddharta Dec 3 '12 at 14:30
I have edited your question slightly. In particular it reads now $(x,y)$ is on the ellipse and not "in" the ellipse. –  Christian Blatter Dec 3 '12 at 14:45
Doing a quick experiment using Cinderella, it seems that your intuition about what minimum point you expect is wrong: there are points off the axes of symmetry where the distance becomes a lot smaller. The larger the excentircity, the more pronounced this effect becomes. –  MvG Dec 3 '12 at 17:30
en.wikipedia.org/wiki/Lagrange_multipliers describes how to maximize $f$ subject to $g=c$. So you'd have $g=c$ as the formula of your ellipsis, and $-D$ as the thing you want to maximize. Now plug these things into the formulas on Wikipedia, compute three derivatives, and you should have a set of three equations. Are you allowed to do this using a computer algebra system? If not, do things become messy enough to actually prevent a manual solution? If so, then re-examine your term above (which I have not checked) and see whether there might be some error in there. –  MvG Dec 3 '12 at 17:39

1 Answer 1

I think it is not necessary to use Lagrange Multipliers. we can replace $\dfrac{x^2}{a^2}$ with $1- \dfrac{y^2}{b^2}$, then we have:

$D(y)=2\dfrac{(\dfrac{1}{a^2}(1- \dfrac{y^2}{b^2})+ \dfrac{y^2}{b^4})^\frac{3}{2}}{\dfrac{1}{a^4}(1- \dfrac{y^2}{b^2})+ \dfrac{y^2}{b^4}}=2\dfrac{(C_1+C_2y^2)^\frac{3}{2}}{C_3+C_4y^2}$, and $C_1=\dfrac{1}{a^2},C_2=\dfrac{1}{b^2}(\dfrac{1}{b^2}-\dfrac{1}{a^2})=\dfrac{a^2-b^2}{a^2b^4}$,$C_3=\dfrac{1}{a^4},C_4=\dfrac{1}{b^2}(\dfrac{1}{b^4}-\dfrac{1}{a^4})=\dfrac{a^4-b^4}{a^4b^6}$, let $z=y^2$,we have a very simple fomula:




since $\dfrac{(C_1+C_2z)^\frac{1}{2}}{C_3+C_4z}>0$, when $D'(z)=0$, we have:

$\dfrac{3C_2}{2}-C_4*\dfrac{C_1+C_2z}{C_3+C_4z}=0$, then we put all staff in, we get: $D'(z)=F(z)*Q(z), F(z)>0$,$Q(z)=b^6-2a^2b^4+(a^4-b^4)z$ , if $D'(z)=0$,then $Q(z)=0$,

$z=\dfrac{b^4(2a^2-b^2)}{a^4-b^4}$, now here is a trick:

$z=y^2 \leq b^2$ then $\dfrac{b^4(2a^2-b^2)}{a^4-b^4} \leq b^2$,ie.$a^2 \geq 2b^2$ it means only under this condition, $D'(z)=0$ can be satisfied.

when $a^2 \geq 2b^2$, we have:

$Q(z)=b^4(b^2-z)+a^2(a^2z-2b^4), $when $z=b^2, Q(z) \geq 0$ ie $D'(z)\geq 0$;,when $ z=0, Q(z)=b^4(b^2-2a^2) < 0$,ie $D'(z)<0$ , which means D(z) has min. put $z=\dfrac{b^4(2a^2-b^2)}{a^4-b^4}$,we get:

$ D_{min}(z)=\dfrac{\sqrt{27}a^2b^2}{(a^2+b^2)^\frac{3}{2}} \leq 2b $

when $a^2 < 2b^2$, we have:

$Q(z)=a^2z(a^2-2b^2)+b^2(2a^2-b^2)(z-b^2)<0$, which mean when $z=b^2$, $D(z)$ has its min which is $2b$.

BTW, this is a Japanese Temple Geometry problem which discussed about 300 years ago.

share|improve this answer
It looks easier to minimize $D^2(x, y)$, since the square roots go away and the resulting $x$ and $y$ will be the same. –  marty cohen Jun 16 '13 at 6:43
You might well think so. I tried that out thinking there'd be a cleaner and quicker way to deal with the algebra. Alas, it is tidier but doesn't reduce the amount of expression-wrangling... –  RecklessReckoner Feb 20 '14 at 8:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.