# Using Lagrange multipliers to find distance from the origin to $4x^2-10xy+4y^2=36$

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:

$$f_x=\lambda g_x, \; f_y=\lambda g_y, \; 4x^2-10xy+4y^2=36$$

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!

• Plug in $y=-x$, then you get $4x^2-10(x)(-x)+4(-x)^2=36$ this gives $8x^2+10x^2=36=18x^2$ which gives $x=+\sqrt(2)$ or $-\sqrt(2)$. Am I missing something? This gives you both $y$ and $x$ since if you take the positive root for $x$, you get the negative root for $y$ and vice versa. So just ignore the complex root. Jan 31 '20 at 16:36

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 $$(x,y)=\left(\pm\sqrt2,\mp\sqrt2\right)$$.