where is my mistake with this equation ot hyperbolic problem?

Question

Let parabola $\Gamma_{1}:$$y^2=4x$,and hyperbolic curve $\Gamma_{2}$: $x^2-y^2=1$.it is well known this two is symmetric.so the two point $A$ and $B$ about $x$ axial symmetry,or mean $x_{A}=x_{B}$.see figure,

but we use $$\begin{cases} y^2=4x\\ x^2-y^2=1 \end{cases}$$ we have $$x^2-4x-1=0\Longrightarrow x=2+\sqrt{5},2-\sqrt{5}$$ so the two point $\color{red}{x_{A}=2+\sqrt{5},x_{B}=2-\sqrt{5}}?$,then contradiction it is well known$x_{A}=x_{B}$,where is mistake?and the second root $\color{blue}{x=2-\sqrt{5}}$ where is from?or where is point to this figure

Thanks help

Answer 1

I know it's rude to give complex solutions to real problems, but I think we can easily see where the second solution $x=2-\sqrt{5}$ came from.

We want to find solutions for: $$\begin{cases} y^2=4x \\ x^2-y^2=1 \end{cases}$$ Let's try find them in $\mathbb{C}$.

We got, as you said before, the equation $x^2-4x-1=0$ that gives us the solutions you found. We will call them $x_1=2+\sqrt{5}$ and $x_2=2-\sqrt{5}$.

But now we have to find $y$. Let's use $x_1$ in the first equation to find $y^2=8+4\sqrt{5}$ which will give us $y_1=\sqrt{8+4\sqrt{5}}$ and $y_2=-\sqrt{8+4\sqrt{5}}$.

You can see that the points you have in the picture are $A=(2+\sqrt{5},\sqrt{8+4\sqrt{5}})$ and $B=(2+\sqrt{5},-\sqrt{8+4\sqrt{5}})$, and that solves part of the contraddiction.

But where does $x_2$ come from? Let us see that $x_2=2-\sqrt{5}<0$ (in fact you noticed $x_1x_2=-1<0$) so the equation $y^2=8-4\sqrt{5}$ has no real solution! Instead in $\mathbb{C}$ we could get $y_3=i\sqrt{8+4\sqrt{5}}$ and $y_4=-i\sqrt{8+4\sqrt{5}}$ which define other two points $C=(2-\sqrt{5},+i\sqrt{8+4\sqrt{5}})$ and $D=(2-\sqrt{5},-i\sqrt{8+4\sqrt{5}})$. These two points are not seen in the picture because they are complex points (they've got imaginary $y$ coordinate) so thay cannot be represented in a picture on the $\mathbb{R}^2$ plane.

In your problem you want to discuss only real intersections of the curves, so you don't notice there are two complex intersections (and you don't want to because you're working with real numbers, so the second equation we solved in $\mathbb{C}$ doesn't give any solution in $\mathbb{R}$). Then the second solution for the equation in $x$ is to be rejected because cannot provide a real solution for the system.

Hope I helped

Answer 2

The confusion arises from ignoring an implicit condition of the first equation, $y^2 = 4x$. Since $y^2 \geq 0$, we have the condition $x \geq 0$.

Thus of the two solutions for $x$ you have, the positive $x = 2 + \sqrt{5}$ is the only solution.

We see that there are two values of $y$ that satisfy the equations with the $x$ value above, and this agrees with the graphical representation.

Answer 3

$x_B$ is an extraneous solution. Try finding the value of $y$ in $y^2=4x_B$ and you'll find nonreal answers.

Answer 4

Ahh your edit clarified your mistake. While $x_A=x_B$ if you are looking at the $x$-coordinates of the two points you have in your picture these $x_A$ and $x_B$ don't correspond to the two solutions of the quadratic you found.

One of those solutions $x=2-\sqrt{5}$ does not solve the pair of equations since assuming we are working over the reals, $x$ must be positive. The two solutions you see in your picture both come from the $x=2+\sqrt{5}$ solution for which you get two different $y$ values when you plug back in to $y^2=4x$ to solve for $y$.