My assignment question is to use Lagrange multipliers to find the points of minimum distance from the origin to the curve $4x^2-10xy+4y^2=36$.
Taking my main function for the purpose of finding critical points as the function of the distance squared, i.e. $f(x,y)=x^2+y^2$ constrained by $g(x,y)=4x^2-10xy+4y^2$, I get the following system of equations to solve:
$$\left\{\begin{array}{l}f_x=\lambda g_x\\f_y=\lambda g_y\\4x^2-10xy+4y^2=36.\end{array}\right.$$
Because the first two equations are symmetrical, $2x=\lambda (8x-10y)$ and $2y = \lambda (8y-10x)$ simplify to $y^2=x^2$ or $y=\pm x$. But plugging the latter into $4x^2-10xy+4y^2=36$ gives me the square root of a negative number. This would appear to show that points of minimum or maximum distance do not exist, but graphing the problem shows that the curve is a set of two hyperbolas which are definitely closest to 0 when $y=\pm x$.
I would really appreciate if someone could explain to me what I might be doing wrong here. I do specifically need to use the Lagrange multipliers method to solve this problem, as per the requirements of the assignment.
Thank you!
You should solve the system$$\left\{\begin{array}{l}2x=\lambda(8x-10y)\\2y=\lambda(-10x+8y)\\4x^2-10xy+4y^2=36,\end{array}\right.$$which is equivalent to$$\left\{\begin{array}{l}(1-4\lambda)x+5\lambda y=0\\5\lambda x+(1-4\lambda)y=0\\2x^2-5xy+2y^2=18.\end{array}\right.\tag1$$The first two equations form a system of two linear equations in the variables $x$ and $y$, depending upon the parameter $\lambda$. If $\lambda$ is such that the only solution is $(0,0)$, forget it; $(0,0)$ is not a solution of the third equation. On the other hand\begin{align}\begin{vmatrix}1-4\lambda&5\lambda\\5\lambda&1-4\lambda\end{vmatrix}=0&\iff-9 \lambda ^2-8 \lambda +1=0\\&\iff\lambda=-1\vee\lambda=\frac19.\end{align}If $\lambda=-1$, then $(1)$ becomes$$\left\{\begin{array}{l}5x-5y=0\\-5x+5y=0\\2x^2-5xy+2y^2=18,\end{array}\right.$$and, yes, this will lead to square roots of negative numbers. But if $\lambda=\frac19$, then $(1)$ becomes$$\left\{\begin{array}{l}\frac59x+\frac59y=0\\\frac59x+\frac59y=0\\2x^2-5xy+2y^2=18,\end{array}\right.$$and now you have the solutions $(x,y)=\left(\pm\sqrt2,\mp\sqrt2\right)$.