# Graphing inequalities

$xy\geq 4$

The answer my book gives me is a hyperbola, but I always thought a hyperbola was a difference between two variables, how does this work?

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BTW, if you have a Mac, then Grapher can plot inequalities. Try also GrafEq, which sadly does not run in current Mac OS X, only in Windows now. – lhf May 3 '11 at 20:56
Does it matter to you why it's a hyperbola or does it only matter why the graph looks the way it does? – Isaac May 3 '11 at 22:20

To graph the region of the plane corresponding to $xy\geq 4$, you start by graphing the region of the plane corresponding to $xy=4$. This is the graph of $y=\frac{4}{x}$, which is a hyperbola (not a parabola).

The region in which $xy\geq 4$ is then: for $x\gt 0$, the region where $y\geq \frac{4}{x}$: the portion of the plane above the right branch of the hyperbola. For $x\lt 0$, the region where $xy\geq 4$ is the same as the region where $y\leq \frac{4}{x}$ (since we are dividing by a negative number), so it is the region below the left branch of the hyperbola.

Added 1. The reason you are allowed to the transformations I did above, dividing by $x$, is that I am being very careful with them. You are correct that you should not simply manipulate the inequality because you run the risk of changing the shape. The things you need to be careful with when doing inequalities are

• Dividing by $0$; this occurs when you divide by an expression involving variables that you do not know whether it can be equal to $0$ or not; this can result in removing valid solutions.
• Multiplying by $0$; same sort of problem.
• Multiplying or dividing by a negative number; because this changes the inequality. This is again an issue if you are dividing/multiplying by expressions involving the variable.
• There are other manipulations that can change the solutions; squaring an equation/inequality can introduce "false positives" (spurious solutions), since $a^2=b^2$ does not imply $a=b$. One has to be careful when multiplying equalities or trying to take square roots.

To take this example, if I were to try to go from $xy\geq 4$, and said "dividing by $x$, we get $y\geq 4/x$," then I would be changing the shape, because I don't know if $x$ is positive or negative or zero, so I don't know if (i) I can actually divide by $x$; that is, if $x\neq 0$; and I don't know (ii) whether dividing by $x$ changes the inequality sign from $\geq$ to $\leq$ (as it would if $x$ is negative).

So you'll notice that I handled the different cases separately: (i) First, we see what happens if $x$ is positive. In that case, dividing by $x$ does not change the inequality, so we are left with $y\geq 4/x$ and $x\gt 0$. That's why we only got the right branch of the hyperbola. (ii) Then, we see what happens if $x$ is negative. In that case, dividing by $x$ does change the inequality, so we are left with $y\leq \frac{4}{x}$ and $x\lt 0$. That's why we only got the left branch of the hyperbola. (iii) Finally, we consider the case where $x=0$; then we have $xy\geq 4$, which is impossible, since $x=0$ means $xy=0$, which is not greater than or equal to $4$.

So by considering the different possibilities carefully and separately we ensure we aren't "changing the shape" when we manipulate the inequality.

The reasons your commented attempt, going from $x^2+y^2=9$ to $x+y=3$ fails, are legion. First, you are taking the square root on the left incorrectly: $$\sqrt{a+b}\neq \sqrt{a} + \sqrt{b}.$$ Lots of students have asserted this since time immemorial; it's still wrong. (Just try it with $a=b=2$; the left hand side is $\sqrt{4}=2$, the right hand side is $2\sqrt{2}\approx 2.82$). Second, even if you did have $\sqrt{x^2}$, you would get $|x|$, not $x$, because you don't know the sign of $x$.

For example, suppose you had $x^2y^2=9$. If you simply take square roots on both sides, $\sqrt{x^2y^2} = \sqrt{9}$, and write this as $xy=3$, then you are in fact "losing" solutions: because $x=-3$, $y=1$ is a solution to $x^2y^2=9$, but not to $xy=3$. That's because $\sqrt{x^2y^2}$ is not equal to $xy$, it's equal to $|xy|$, the absolute value of $xy$. In general, $\sqrt{a^2}=|a|$, not $a$.

You'll notice, I trust, how careful I was being with the manipulations I did do to $xy\geq 4$ to ensure I wasn't changing the equation.

Added 2. Why is $xy=4$ a hyperbola?

A general conic (circle, ellipse, parabola, hyperbola, or degenerate conic) is given by a formula of the form $$ax^2 + bxy + cy^2 + dx + ey + f = 0$$ where $a,b,c,d,e,f$ are constants, and at least one of $a$, $b$, and $c$ are not zero. Under a suitable change-of-variable, which amounts to translating and/or rotating the conic so that its center (in the case of circles, ellipses, and hyperbolas) or its vertex (in the case of parabolas) lies at the origin, and its axes are horizontal and vertical (in the case of ellipses and hyperbolas) or its directrix is horizontal or vertical (in the case of parabolas); (with circles, the symmetry means it doesn't matter).

By doing suitable translations and rotation, we can always bring the equation into a much nicer form, namely something of the form $$\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$$ for circles and ellipses (it's a circle if $A^2=B^2$, an ellipse otherwise); of the form $$\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1\qquad\text{or}\qquad \frac{y^2}{B^2}-\frac{x^2}{A^2} = 1$$ for hyperbolas; and of the form $$y = Ax^2\qquad\text{or}\qquad x = By^2$$ for parabolas.

(There are other more general forms if you want to allow the center to be somewhere other than the origin; see for example the Wikipedia page on conic sections.)

In this case, the change of variable \begin{align*} x &= \frac{\sqrt{2}}{2}x' + \frac{\sqrt{2}}{2}y'\\ y&= -\frac{\sqrt{2}}{2}x' + \frac{\sqrt{2}}{2}y', \end{align*} which corresponds to rotation the plane by $45$ degrees, with the $x'=0$ line corresponding to the line $x=y$ and the new axis $y'=0$ to the line $x=-y$ (the scary looking fractions $\frac{\sqrt{2}}{2}$ comes from the fact that $\sin(45^{\circ}) = \cos(45^{\circ}) = \frac{\sqrt{2}}{2}$),
gives that $xy=4$ is equivalent to $$4 = xy = \left(\frac{\sqrt{2}}{2}x' + \frac{\sqrt{2}}{2}y'\right)\left(-\frac{-\sqrt{2}}{2}x' + \frac{\sqrt{2}}{2}y'\right) = \frac{1}{2}y'^2 - \frac{1}{2}x'^2$$ so we are looking at $$\frac{1}{8}y'^2 - \frac{1}{8}x'^2 = 1.$$ If you draw the lines $x=y$ and call it the $x'$-axis; and $x=-y$ and call it the $y'$-axis, then the graph of $xy=4$ is the same as the graph of the hyperbola below. The change of variables, though, is not very good for discussing inequalities (you can do it, but it makes things difficult; better to stay with $xy$-coordinates). This is just to show you why $xy=4$ is "really" a hyperbola, even though you are not seeing a difference of squares.

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@Adam. A number of people have put a good amount of time into answering your questionS. The least you can do is accept an answer, and refrain from using profanity here.]It's completely inappropriate to ask a question, and then get angry with people for not giving you the answer you'd like. We're pointing out errors in your work; do you want us to pat you on the head, and tell you "good work"...when you're clearly making mistakes? – amWhy May 3 '11 at 21:14
@Adam: The main problem is that you think that there are "thousands and thousands of rules". In fact, there are very few that are logically connected to one another, but by trying to figure them out in isolation, you face a very difficult task. Also, you are not thinking throught things carefully. In fact, $xy=4$ is of the form $ax^2+bxy+cy^2+dx+ey+f=0$: you have $a=c=d=e=0$, $b=1$, and $f=-4$, which gives $xy-4=0$, which is exactly the same as $xy=4$. Note that with that choice of coefficients, $a,b,c$ are not all zero. – Arturo Magidin May 3 '11 at 21:29
@Adam: the last sentence of your last comment was certainly uncalled for. Arturo here has devoted a good chunk of his time in guiding you to understanding; I'm pretty sure an "elitist" would not have bothered with you. Your attitude as currently set (there certainly aren't "thousands of rules"; remember a few and you can often deduce the others) is hindering you here... – J. M. May 4 '11 at 0:53
@Adam: To quote Jon Stewart: "I didn't call your sister a prostitute, I just said she takes money in exchange for having sex with men she doesn't know." – Arturo Magidin May 4 '11 at 3:52
@Adam: But keep in mind: you are talking to mathematicians (of some level or another); insulting the field to which we dedicate ourselves (to some degree or another) is not exactly singing our praises, even if it is not insulting us personally. You wouldn't go to the locker room of the L.A. Dodgers and start saying how much you hate baseball and how stupid all the rules you have to memorize are, would you? – Arturo Magidin May 4 '11 at 4:52

I think you are probably used to seeing the equation of hyperbola being written as $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ However, note that if we transform variables as $\tilde{x} = \frac{x}{a} + \frac{y}{b}$ and $\tilde{y} = \frac{x}{a} - \frac{y}{b}$, then the equation transforms as $$\tilde{x} \tilde{y} = 1$$ Note that the map from $(x,y) \rightarrow (\tilde{x},\tilde{y})$ is a nice bijective map

Essentially what we are doing is to view the hyperbola from a different co-ordinate system

In case of your inequality, we have $$xy \geq 4$$

If we transform this using $\tilde{x} = \frac{x+y}{2}$ and $\tilde{y} = \frac{x-y}{2}$, we get $x = \tilde{x} + \tilde{y}$ and $y = \tilde{x} - \tilde{y}$ and the equation in $\tilde{x}$ and $\tilde{y}$ become $$\tilde{x}^2 - \tilde{y}^2 \geq 4$$

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These pictures were made with Grapher, which I mentioned in my comment to the question. – lhf May 3 '11 at 21:05