Lagrange's Multiplier What is the distance of the farthest point lying on the curve $2x^2 ＋ 6xy ＋ 5y^2 ＝1$ from the origin? It can be solved by polar coordinates but I was trying to do it by Lagrange multipliers by this method $x^2＋y^2+ c(2x^2 ＋ 6xy ＋ 5y^2 -1)$, but I am not getting any maximum answer from it.
 A: Work with the square of the distance, that is, let $f(x,y)=x^2+y^2$. So, you get the system$$\left\{\begin{array}{l}2x=\lambda(4x+6y)\\2y=\lambda(6x+10y)\\2x^2+6xy+10y^2=1\end{array}\right.$$The first two equation form a system of linear equations dependent upon $\lambda$. So, choose $\lambda$ such that the determinant of the matrix of the coefficients of the system is $0$; otherwise, the only solution will be $(0,0)$, which is not a solution of the third equation.
A: Hint: use the formula $$d=\sqrt{x^2+y^2}$$ where $$2x^2+6xy+5y^2=1$$
Use the function
$$F(x.y.\lambda)=\sqrt{x^2+y^2}+\lambda(2x^2+6xy+5y^2-1)$$
and differentiate with respect to $x,y,\lambda$
or you plug $$y=\frac{1}{5}(-3x-\sqrt{5-x^2})$$
or $$y=\frac{1}{5}(-3x+\sqrt{5-x^2})$$
in your objective function and you will get a Problem in only one variable.
A: $$2x^2+6xy+5y^2=1$$ is the graph of an ellipse centered at the origin with the vertices along the eigenvectors of the matrix, $$ A=\begin{pmatrix}2&3\\3&5\end {pmatrix}$$.
The eigenvalues of $A$ are  $$\frac {7\pm \sqrt 45}{2}$$
Thus the maximum distance from the origin is $$(\frac {2}{   7- \sqrt 45})^2 \approx 46.97$$
