Equation system to solve

Question

I am trying to solve the following system of equations:

$$i) \space 6x-y+\lambda2x=0 \\ii)\space2y-x+\lambda2y=0\\iii)\space x^2+y^2-1=0$$

It seems like that no single line contains a solution on its own, except 0 which is not fitting the third line. I tried to do the following:

$$ i*y-ii*x=4xy-y^2-x^2$$I solved this equation by abc-formula and received $$-4y +\sqrt{3},\space-4y -\sqrt{3}$$

But I am not sure what to do with this solution.

So my question is of course not to solve this for me - but do you have any recommendations to solve? Or another method that may solve problems like this?

Answer 1

Think geometrically. You want points common to two lines through the origin which also cut the circle $x^{2}+y^{2}=1$ in a common point. This is possible only when the two lines are identical. Can you use this information to answer the question?

Answer 2

HINT

I suppose you are solving a problem by Lagrange's multipliers, then we have from the "$i$"

• $y=6x+2\lambda x$

and from the "$ii$"

• $2(6x+2\lambda x)-x+2\lambda(6x+\lambda2x)=(4\lambda^2+16\lambda+11)x=0$

that is

• $x=0 \implies y=0$ not acceptable
• $4\lambda^2+16\lambda+11\implies \lambda=-2\pm \frac{\sqrt 5}2$

As an alternative note that the system "i" and "ii" is homogeneous in the form

$$\begin{bmatrix}6+2\lambda&-1\\-1&2+2\lambda\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$$

which has not trivial solution if and only if

$$\det\begin{bmatrix}6+2\lambda&-1\\-1&2+2\lambda\end{bmatrix}=4\lambda^2+16\lambda+11=0$$

which corresponds to the same condition already found.

From here we can find the lines equations $y=mx$ solution of "$i$" and "$ii$" and find the coordinates which satisfy "$iii$".

Answer 3

The first two equations are homogeneous, and by setting $z:=y/x$,

$$\begin{cases}6-z+2\lambda=0,\\2z-1+2\lambda z=0\end{cases}.$$

You can eliminate $\lambda$ (which you presumably don't need), giving

$$z^2-4z-1=0.$$

Solve, and then from the last equation,

$$x=\pm\frac1{\sqrt{1+z^2}}, y=zx.$$

Answer 4

From the first and second equations we get :

$\lambda=\frac{y-6x}{2x}=\frac{x-2y}{2y}$;

this means that

$y=x(2-\sqrt{5})$,

$y=x(2+\sqrt{5})$;

replacing the values of $y$ in the third equation we get:

$x=-\sqrt{\frac{1}{2}-\frac{\sqrt{5}}{5}}$,

$x=+\sqrt{\frac{1}{2}-\frac{\sqrt{5}}{5}}$.

The system solutions are four $(x,y,\lambda)$ of values:

$(\sqrt{\frac{\sqrt{5}}{5}+\frac{1}{2}}, -\sqrt{-\frac{\sqrt{5}}{5}+\frac{1}{2}}, -2-\frac{\sqrt{5}}{2})$;

$(-\sqrt{\frac{\sqrt{5}}{5}+\frac{1}{2}}, +\sqrt{-\frac{\sqrt{5}}{5}+\frac{1}{2}}, -2-\frac{\sqrt{5}}{2})$;

$(\sqrt{-\frac{\sqrt{5}}{5}+\frac{1}{2}}, +\sqrt{+\frac{\sqrt{5}}{5}+\frac{1}{2}}, -2+\frac{\sqrt{5}}{2})$;

$(-\sqrt{-\frac{\sqrt{5}}{5}+\frac{1}{2}}, -\sqrt{+\frac{\sqrt{5}}{5}+\frac{1}{2}}, -2+\frac{\sqrt{5}}{2})$.