# How to find what are the points closest to and farthest from (0,0) of ellipse $9x^2+4y^2=36$ using optimization?

Please do not use Lagrange multipliers. Assume these have not been introduced and optimize.

Edit: I try optimizing the squared distance formula using the equation as a constraint, but I only get one critical X out, so I don't know if that is the max or min, or how to find the other, since it's asking for both.

Edit 2: This is a serious question. Please do not discriminate because I cannot get the editor to work.

I get $x=-3/2$ and two corresponding y values, but is this the max or the min? Where do I plug this in to find out?

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Can you share what you've tried, and explain what's giving you trouble? Do you know any techniques other than Lagrange multipliers? – user61527 Jun 18 '14 at 6:56
@conjuggersz do you mean 9 x^2 + 4 y^2 = 36? – bobbym Jun 18 '14 at 7:09
Please check the formula in your title. Should it be $4x^2+9y^2=36$? – mike Jun 18 '14 at 7:10
Why don't you draw the ellipse ? – Claude Leibovici Jun 18 '14 at 7:21
See edits. Title corrected. I don't know enough LaTex to type out what I get, but I end up with only one critical point for x. The method I am trying to practice is optimizing and finding the points algebraically – conjuggersz Jun 18 '14 at 7:36

This problem can be done without any calculations whatsoever just by understanding the graph of an ellipse in standard form. However just for fun I provide a calculation using calculus without the use of any transcendental functions.

We are trying to find the points on the ellipse, $\frac{x^2}{4}+\frac{y^2}{9} = 1$, which are closest to and farthest away from the origin. This means that we must optimize the expression $d^2 = x^2+y^2$ subject to the given constraint. To do this we first introduce the parameterization,

$$x = 2 \cdot \frac{2t}{1+t^2} \qquad y=3 \cdot \frac{1-t^2}{1+t^2}.$$

Substituting our parameterization into the expression for $d^2$ we get,

$$d^2 = \frac{16 t^2 + 9(1-t^2)^2}{(1+t^2)^2}.$$

Now we look for critical points by examining $(d^2)^\prime$,

$$(d^2)' = \frac{d}{dt} \frac{16 t^2 + 9(1-t^2)^2}{(1+t^2)^2} = \frac{40(t)(t^2-1)}{(t^2+1)^3},$$

the above expression is equal to zero when $t=0,1,-1$ or $t=\pm \infty$. To determine which values correspond to the minimum and maximum respectively we simply evaluate $x$ and $y$ for all the different values of $t$. We will manually pick out the values which are closest to and furthest from the origin.

$$x(0) = 0 \quad y(0)=3$$ $$x(1) = 2 \quad y(1)=0$$ $$x(-1) = -2 \quad y(-1)=0$$ $$x(\infty) = 0 \quad y(\infty) = -3$$ $$x(-\infty) = 0 \quad y(-\infty) = -3$$

The points furthest from the origin are $(0,3)$ and $(0,-3)$. The points closest to the origin are $(2,0)$ and $(-2,0)$.

Edit In response to the comment:

$t$ is just a parameter which labels the points on the ellipse. To understand the validity of this parametrization note that,

$$\frac{1}{4} \left( 2 \cdot \frac{2t}{1+t^2} \right)^2 + \frac{1}{9} \left( 3 \cdot \frac{1-t^2}{1+t^2}\right)^2 =1.$$

For a proof just google "rational points on the circle".

Its not necessary to do it this way I just happen to enjoy this rational parametrization. What you are suggesting about substituting $y^2$ as a function of $x$ into the $d^2$ equation will work just as well. Also you could parameterize with $x=2\sin \theta$ and $y=3\cos \theta$.

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I get this but what does the t represent? I have just substituted the y^2 value in given the distance equation and derived it in terms of x, then set equal to zero. I don't understand your parameterization step – conjuggersz Jun 20 '14 at 4:03

Note that the distance from the centre, which you want to measure is given by $r^2=x^2+y^2$ with $r\ge 0$ and maximising/minimising $r$ is therefore equivalent to maximising/minimising $r^2$.

To maximise $r^2$ note that $5x^2+4r^2=36$ so that $4r^2\le 36$ with equality, and a therefore a maximum, only if $x=0$.

To minimise $r^2$ note that $9r^2-5y^2=36$ so that $9r^2\ge 36$ with equality, and a minimum, only if $y=0$.

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