Maxima/Minima Problem

Question

I am working on this Maxima and Minima Problem :

Determine the Max and Min distance of origin from the curve $3x^2+4xy+6y^2=140$

I tried it solving using the lagrange's method of multipliers. I get the following equations

$x+3x\theta+2y\theta=0$

$y+4y\theta+2x\theta=0$

$2z=0$

$3x^2+4xy+6y^2=140$

So i get $z=0$

Please suggest how to go about solving for $x$ and $y$ ?

Answer 1

Another approach. Draw a circle centered at the origin. You want the radii which intersect the ellipse in exactly two points. So apply the equation $x^2+y^2=r^2$. Define $A=\frac{140}{r^2}$. Then $3x^2+4xy+6x^2=A(x^2+y^2)$. Divide through by $x^2$, and define $z=\frac{y}{x}$, yielding a quadratic in z. We want the discriminat to equal zero, which occurs when $A=7$ or $2$. Thus $r^2=20$ or $70$.

Answer 2

\begin{align} \min_{x,y}&x^2+y^2\\s.t.\\ 3x^2+6y^2+4xy&=140\\ L(x,\lambda)&=x^2+y^2-\lambda(3x^2+6y^2+4xy-140)\\ \nabla L(x,\lambda)&=\binom{2x}{2y}-\lambda\binom{6x+4y}{12y+4x}=0\\ \text{Necessary Conditions:}\\ 2x-\lambda(6x+4y)&=0\\ 2y-\lambda(12y+4x)&=0\\ 3x^2+6y^2+4xy&=140\\ \end{align} Solve these 3 equations and you have your answer. \begin{align} 2x&=\lambda(6x+4y)\\ \lambda&=\frac{2x}{6x+4y}\\ 2y-\lambda(12y+4x)&=0\\ 2y-\frac{2x}{6x+4y}(12y+4x)&=0\\ y&=\frac{x}{6x+4y}(12y+4x)\\ (6x+4y)y&=x(12y+4x)\\ 6xy+4y^2&=12xy+4x^2\\ 2y^2&=3xy+2x^2\\ 3x^2+6y^2+4xy&=140\\ \text{Solving you get,}\\ x=\pm2,y=\pm4\\ d=20 \end{align}

Answer 3

A third approach. Rotate the ellipse to bring the major and minor axes coincident with the coordinate axes. Then the equation becomes $\lambda_1x^2+\lambda_2y^2=140$, where $\lambda_1$ and $\lambda_2$ are the eigenvalues of the symmetric matrix which generates the quadratic form on the left side of the equation. In this case the eigenvalues are 7 and 2. It is not necessary to calculate the eigenvectors.