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Find the points on the ellipse $2x^2 + 4xy + 5y^2 = 30$ closest and farthest from origin. How to do this problem? I know how to find a closest point if $z = f(x,y)$ is given, however, this is 2 dimensional.

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4 Answers 4

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You can use Lagrange multipliers, too, since you have to find the minimum/maximum of the function $f(x,y)=x^2+y^2$ with the constraint $2x^2+4xy+5y^2-30=0$. We get:

$$\frac{2x+4y}{x}=\frac{4x+5y}{y},$$

that is equivalent to:

$$\frac{y}{x}-\frac{x}{y}=\frac{3}{4},$$

so, in order to find the closest/farthest point from the origin, you simply have to intersect the ellipse with the two lines $y=mx$ for

$$ m = \frac{y}{x} = \frac{3\pm\sqrt{73}}{8}. $$

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If Lagrange multipliers are not yet part of the toolkit, or even if they are, we can use a perpendicularity argument.

Visualize circles $x^2+y^2=r^2$, where $r$ is a variable that starts very close to $0$, and increases.

There is a *smallest*value of $r$ such that $x^2+y^2=r^2$ meets our ellipse, and a biggest. These are what we are looking for. At these values of $r$, the circle $x^2+y^2=r^2$ is tangent to our ellipse.

Let the point of tangency be $P=(a,b)$. Then $$2a^2+4ab+5b^2=30.\tag{$1$}$$ The line $OP$ is perpendicular to the tangent at $(a,b)$ of the circle, and hence of the ellipse.

Now we can use either one variable calculus (implicit differentiation) or partial derivatives. If $f(x,y)=2x^2+4xy+5y^2-30$, we have $f_x(x,y)=4x+4y$ and $f_y(x,y)=4x+10y$.

The perpendicularity condition is that the dot product of $(4a+4b, 4a+10b)$ and $(a,b)$ is $0$. So we get $$a(4a+4b)+b(4a+10b)=4a^2+14ab+4b^2=0.\tag{$2$}$$

Solve, using the Qadratic Formula. We get $$a=\left(\frac{-7\pm\sqrt{33}}{4}\right)b.$$ Now substitution in Equation $(1)$gives our candidates.

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Hint: Use Lagrange multipliers!

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I'm not familiar with that? Could you be more specific? Thanks –  Frank Xu Oct 15 '12 at 4:35

Another way is use Rotation of axes, to eliminate the $xy$ term.

Here $\cot 2\theta=\frac{2-5}4=-\frac 3 4$

$\frac {\cos 2\theta}3=\frac{\sin 2\theta}{-4}=\frac 1{\pm 5}$ (Using squaring & adding)

If $\cos 2\theta=\frac 3 5,\sin 2\theta=-\frac 4 5$

Using $\cos 2\theta=2\cos^2\theta-1=1-2\sin^2\theta$, we get $\sin^2\theta=\frac 1 5$ and $\cos^2\theta=\frac 4 5\implies \cos\theta=\pm\frac 2{\sqrt 5},\sin\theta=∓\frac 1{\sqrt 5}$ as $\sin 2\theta<0$, the sign of $\cos\theta,\sin\theta$ will be opposite.

The equation of the ellipse in the rotated axes becomes, $$\frac{x'^2}{30}+\frac{y'^2}{5}=1$$

Now in orthogonal transformation, the distance between two points is one of the invariants and in rotation, the origin remains unchanged and any point of this curve can be P$(\sqrt{30}\cos \phi,\sqrt5 \sin \phi)$.

The distance of P from the origin is $\sqrt{30\cos^2\phi+5 \sin^2 \phi}$ $=\sqrt{30-25\sin^2 \phi}$

The distance will be minimum $(=\sqrt 5)$ if $\sin^2 \phi=1$ i.e., from $(0,\pm\sqrt 5)$ in the rotated co-ordinates.

The distance will be maximum $(=\sqrt {30})$ if $\sin^2 \phi=0$ i.e., from $(\pm\sqrt {30},0)$ in the rotated co-ordinates.

Now use $x=x'\cos\theta-y'\sin\theta$ and $y=x'\sin\theta+y'\cos\theta$.

We have already derived, $\cos\theta=\pm\frac 2{\sqrt 5},\sin\theta=∓\frac 1{\sqrt 5}$

For the minimum distance, if $\cos\theta=\frac 2{\sqrt 5},\sin\theta=-\frac 1{\sqrt 5}$

If $y'=\sqrt 5,x=-\sqrt 5(-\frac 1{\sqrt 5})=1$ and $y=\sqrt5(\frac 2{\sqrt 5})=2$

Similarly, if $y'=-\sqrt 5, (x,y)$ will be $(-1,-2)$

The maximum case can be handled in the same way.

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