Just taking (failing) a simple algebra class, can't figure this one out and no one can explain it to me and the book just tells me to do it.

Find an equation for the hyperbola described:

foci at $(-4,0)$ and $(4,0)$; asymptote the line $y=-x$.

So I know that since the numbers on the $x$ axis are changing it will be a horizontal hyperbola. That means $0$ is the center and $c$ is $4$.

I know the slope is $b/a$ for horizontal equations so I know that $b/a = -1$

From that I can get $b = -a$.

This is as far as I can get, my book basically does these steps in the solution manual except they get $-b/a = -1$, $b=a$ I don't even know why. I can't work past this point without graphing and I know there is suppose to be a way just by working out the algebra but I don't see a solution.

I might not be prepared for this test but I am prepared to fail the test.

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    $\begingroup$ Clearly, you have some "general form" for the equation of a hyperbola, since you are talking about "a", "b", and "c". What is the general form you have? $\endgroup$ – Arturo Magidin Apr 8 '11 at 16:00
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    $\begingroup$ The only information I am given is foci at (−4,0) and (4,0); asymptote the line y=−x. $\endgroup$ – user9301 Apr 8 '11 at 16:03
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    $\begingroup$ @Glen: For this problem. But you know, in general, what equations for "horizontal hyperbolas" look like, don't you? For instance, that they can be written as $$\frac{(x-A)^2}{a^2}-\frac{(y-B)^2}{b^2} = 1$$for some $A$, $B$, $a$, $b$, or some such general thing about hyperbolas? I mean: you immediately jump to say that "c is 4". What is "c"? Then you talk about "b/a". What are b and a relative to the hyperbola? Obviously you have more information about hyperbolas in general. $\endgroup$ – Arturo Magidin Apr 8 '11 at 16:06
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    $\begingroup$ @Arturo Magidin Not exactly sure what you are asking I just know that (x^2)/a - (y^2)/b = 1 is the forumla and that the formula of the slope works in a way that b/a is going to be equal to the slope. $\endgroup$ – user9301 Apr 8 '11 at 16:08
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    $\begingroup$ @Glen: The "standard" East-West opening hyperbola with center the origin has equation $x^2/a^2 -y^2/b^2=1$. The foci are at $(\pm c,0)$, where $c=\sqrt{a^2+b^2}$. From the comments above, you learned that $a=b$. So $4=\sqrt{2a^2}$. Now you know $a$. $\endgroup$ – André Nicolas Apr 8 '11 at 16:32

So, you know that the equation is going to look like $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1.$$ You already figured out that $a=-b$. So now you need to figure out what values of $a$ and $b$ will make the foci be at the given point $(4,0)$ and $(-4,0)$.

Since you know that $a=-b$, then you know that $a^2=b^2$, so you can rewrite the equation as $$\frac{x^2}{a^2} - \frac{y^2}{a^2} = 1$$ or, by clearing denominators (multiplying through by $a^2$) $$x^2 - y^2 = a^2.$$ Now: where does the hyperbola intersect the $x$ axis? If you have a point $(X,0)$ in the hyperbola, then plugging it in you will have $$X^2 = a^2.$$ So if you can determine where the hyperbola intersects the $x$-axis, you can use that to figure out the value of $a$, and so the value of $b$, and so the equation for the hyperbola.

Note: There are many, many, many ways to finish the problem, depending on what you know about hyperbolas in general (which is precisely why I asked you in comments what you knew).

Added. One possible way to finish the problem is to use the eccentricity of the hyperbola. The distance from the center to the foci is $a\varepsilon$, where $\varepsilon$ is the eccentricity of the hyperbola. The eccentricity is always equal to $$\varepsilon = \sqrt{ 1 + \frac{b^2}{a^2}}$$ so here, since $a=b$, that means that the eccentricity is $\sqrt{2}$. Therefore, since the distance from the center to the foci is $4$, this tells you that $4 = a\varepsilon = a\sqrt{2}$. You now know what $a$ is, and thus, you know what $a^2=b^2$ is, and so you know the equation of the hyperbola.

Added 2. Another possibility you discuss in the comments. You have $a$ is the distance from the center to the vertices; $b$ is the distance from the center to the conjugate axis. These distances satisfy $c^2 = a^2+b^2$, where $c$ is the value $4$ that you have. Since $a^2=b^2$, then you have $c^2=2a^2$, or $c = \sqrt{2a^2} = \sqrt{2}|a| = a\sqrt{2}$. You already know what $c$ is, so now you can solve for $a$. Once you know $a$, you also know $b$ (since $a=-b$) which gives you the information you need.

  • $\begingroup$ Mind blown! I have never seen anything like that with all these circle, ellipse, parabola and hyperbola problems. Most likely the test will all be stuff like that. thank you Actualyl I still can't figure this one out at all. I just can't figure anything out. $\endgroup$ – user9301 Apr 8 '11 at 16:19
  • $\begingroup$ Sorry this is just incredibly frustrating for me that I can't figure any of this stuff out on my own. This is why I fail all my tests I guess, I can't figure out the hard problems without help and that is all the test is. I have no idea what e is and we never used that in class, my book just does the transverse axis in the x asix, c=4 using the slop of the asymptote -b/a = -1 = b=a b^2+b^2 = 16 b=square root of 8 a= square root of 8 $\endgroup$ – user9301 Apr 8 '11 at 16:36
  • $\begingroup$ There are multiple ways of getting the missing information; eccentricity is one of them. You can also use other information, but what you write is not correct: you should not use the = sign between the equation that says $-b/a=1$ and the equation that says $b=a$. I'll add your way. $\endgroup$ – Arturo Magidin Apr 8 '11 at 16:42
  • $\begingroup$ @Glen: I edited your comment to remove the possibly offensive language. $\endgroup$ – Willie Wong Apr 8 '11 at 17:52
  • $\begingroup$ @Willie: It doesn't particularly offend me, but I do notice that the site is kept pretty "clean", which is why I mentined it. $\endgroup$ – Arturo Magidin Apr 8 '11 at 18:08

As the foci are on the $x$-axis and symmetric with respect to the $y$-axis, the equation of the hyperbola is


Such an hyperbola has two asymptotes:

$$y=\frac{b}{a}x\qquad\text{and}\qquad y=-\frac{b}{a}x,$$ where both $a$ and $b$ are positive. The given asymptote is $y=-x$ and the other one is $y=x$. The equation $y=-\frac{b}{a}x$ should be equivalent to $ y=-x$, which implies that $-\frac{b}{a}=-1$. Hence $b=a$. Now you use the information on the foci to find $a$. The distance from each focus to the origin is $c=4$. These numbers are related by the equation $$a^{2}+b^{2}=c^{2} .$$ For $b=a$ and $c=4$, we have $a^{2}+a^{2}=16$, thus $a=2\sqrt{2}$ (the other solution is negative), and the equation of the hyperbola is


  • $\begingroup$ érico: My turn: "axis" and "hyperbola". (-: (Hyperbole is an extravagant exaggeration...) $\endgroup$ – Arturo Magidin Apr 8 '11 at 18:14
  • $\begingroup$ @Arturo: thanks for correcting me. $\endgroup$ – Américo Tavares Apr 8 '11 at 22:35

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