Hyperbolas

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shawie

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Aug 31, 2005
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X^2/a^2 - y^2/b^2 =1 y=+-b/a
Foci c^2=a^2b^2

I know that for the equation Y^2/16 - x^2/4=1 that a = 4; b=2; and c=square root of 13 ...and I know how to graph it and all.

What I don't know how to do is when the equation doesn't equal 1 and the x and y's have coefficients in front of them for example:

25x^2-16y^2=400 ...how do I find a, b, and c?

Another problem that I'm having trouble with is when I'm given a vertex at (0,-12) and a focus at (0,-13). How do I find an equation of the hyperbola with its center at the origin?
 
Hello, shawie!

You know quite a bit about hyperbolas.

\(\displaystyle \L\frac{x^2}{a^2}\,-\,\frac{y^2}{b^2}\:=\:1,\;\;y\,=\,\pm\frac{b}{a}\;\;\)\(\displaystyle \text{Foci: }\,c^2\:=\:a^2\,+\,b^2\)

I know that for the equation: \(\displaystyle \frac{y^2}{9}\,-\,\frac{x^2}{4}\:=\:1\), that \(\displaystyle a\,=\,3,\,b\,=\,2,\,c\,=\,\sqrt{13}\)
and I know how to graph it and all.

What I don't know how to do is when the equation doesn't equal 1 and the x and y's have coefficients in front of them.

For example: \(\displaystyle \,25x^2\,-\,16y^2\:=\:400\)
. . . how do I find \(\displaystyle a,\,b,\) and \(\displaystyle c\) ?
Click to expand...
You'll kick yourself . . .

Divide by 400: \(\displaystyle \L\:\frac{25x^2}{400}\,-\,\frac{16y^2}{400}\:=\:\frac{400}{400}\;\;\)The right side is supposed to be 1, remember?

And we have: \(\displaystyle \L\:\frac{x^2}{16}\,-\,\frac{y^2}{25}\:=\:1\)


I'm given a vertex at (0,-12) and a focus at (0,-13).
How do I find an equation of the hyperbola with its center at the origin?
Click to expand...
The vertices and foci are on the \(\displaystyle y\)-axis . . . it's a "vertical" hyperbola
\(\displaystyle \;\;\)with the form: \(\displaystyle \:\frac{y^2}{a^2}\,-\,\frac{x^2}{b^2}\:=\:1\)

We know that: \(\displaystyle \,b\,=\,-12\) and \(\displaystyle c\,=\,-13\)
And we have the formula: \(\displaystyle \,c^2\:=\:a^2\,+\,b^2\)
\(\displaystyle \;\;\)So we have: \(\displaystyle \,(-13)^2\:=\:(-12)^2\,+\,b^2\;\;\Rightarrow\;\;b^2\,=\,25\)

Therefore, the equation is: \(\displaystyle \L\,\frac{y^2}{144}\,-\,\frac{x^2}{25}\:=\:1\)
 
*kicks herself* see! i should be more confident sometimes... alright cool thanks for clarifying those you guys :)
 
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