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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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    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?2011-04-08
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    The only information I am given is foci at (−4,0) and (4,0); asymptote the line y=−x.2011-04-08
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    @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*.2011-04-08
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    @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.2011-04-08
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    I know that a is the distance from the center to the vertci or in this problem h+/-a and that b is the distance from the center to the conjugate axis and that c is center to the focus.2011-04-08
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    @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 $a$.2011-04-08
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    @Glen: the solution manual gets $a=b$ because this hyperbola has two asymptotes: $y=\frac{b}{a}x$ and $y=-\frac{b}{a}x$, where both $a$ and $b$ are positive. The given asymptote is $y=-x$ and the other one is $y=x$.2011-04-08

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