Two different hyperbolae from the same two linear equations

Question

Suppose two lines are represented by $x-2y=2$ and $x+2y=2$. If I multiply the equations I get $$x^2-4y^2=4$$ which is the equation of a hyperbola. But if I take $x-2y-2=0$ and $x+2y-2=0$ and multiply, I get $$x^2-4y^2-4x+4=0$$ which is another hyperbola (a degenerate one with two crossing lines). Why am I getting two different hyperbolae?

Answer 1

First of all, Only one of the equation represent Hyperbola i.e. $$ x^2-4y^2=4 \\ or \\ \frac{x^2}{4} - \frac{y^2}{1} = 1 $$ represent hyperbola but equation $$ x^2-4y^2-4x+4=0 \\ or \\ \frac{(x-2)^2}{1} - \frac{y^2}{1/4} = 0$$ does not represent hyperbola but only represent set of two lines.

See Figure : ( Green indicates First Equation and Red indicates second )

and the conflict is actually due to zero as, $$ x^2-4y^2-4x+4=0 \\ (x-2y-2)(x+2y-2)=0$$ For equation to be true only one part or factor need to be true which give rise to a line.

(For any pair of x and y that satisfy equation their is no need to satisfy both parts at the same time only one part can lead to satify the equation )

but in this - $$x^2-4y^2=4$$ the whole equation need to be satisfied by pair of x and y as it can not be spitted into such parts.

Answer 2

Why do you expect them to be the same?

In the first case you're doing the multiplication $2\times 2=4$ and in the second $0\times 0=0$. You wouldn't be surprised to see that these products aren't equal, would you?

Answer 3

Okay, so if we assume that the equations you are starting with belong together, x=2 and y=0 (as stated before). Now you have the equation $x^2-4y^2=4$ so you can start and reshape them by doing the following steps: $$x^2-4y^2=4 \qquad| -4 \\ \Leftrightarrow x^2-4y^2-4=0 \\ \Leftrightarrow x^2-4y^2-8+4=0 \\ \Leftrightarrow x^2-4y^2-(4 \cdot 2)+4=0$$

Since we now that x=2, we can also replace it: $x^2-4y^2-4x+4=0$

It has already been said, but those are no hyperbolaes but equations with exactly one solution, that is why you can replace x for 2.

Answer 4

Why do you expect to get the same hyperbola? And more generally, why do you expect to get anything meaningful from multiplying sides of two equations?

Take these two simplest linear equations: $$\begin{cases}x=0 \\ y=0\end{cases}$$ They describe two lines, the $Y$-axis and the $X$-axis, respectively.

If you multiply them as you did to you equations, you'll get $$xy=0$$ which is an equation of both axes (a degenerate case of hyperbola).

However, if you swap sides of one equation, you'll get an equivalent system of equations: $$\begin{cases}x=0 \\ 0=y\end{cases}$$ but the result of multiplying is no longer equivalent to the previous one: $$0=0$$ — you get an equation of the whole plane!

See another modification: let's add a unit to both sides of the first equation: $$ {\begin{cases}x+1=1 \\ y=0\end{cases} \over (x+1)y=0} \quad {\begin{cases}1=x+1 \\ y=0\end{cases} \over y=0} $$ — depending on the sides' order we get either an equation of two lines of an equation of just one line.