hi. i forgot to find the verticies of the hyperbolas. i really feel stupid
(Y-3)^2 /25 - (X-2)^2 /16
(\__/) I became insane, with long intervals of horrible
(='.'=) sanity
(")_(")
lol wut?

lols i said lol WAT
Last edited by dean.guitar.man at May 27, 2008,
well, the denominators give you the axes of the hyperbola so in this case the major axis would be parallel to the y-axis and have a length of ten, the minor parallel to x with a length of 8 (I'm pretty sure). Then the vertices would be at y=3, -3 along the major axis, I believe
^ what he said
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Why 8?
im my book its givin me 2,8 and 2,-2
(\__/) I became insane, with long intervals of horrible
(='.'=) sanity
(")_(")
lol wut
i remember this stuff from from like 3-4 years ago in high school and haven't seen it since.
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I don't know, I'm not much of a math person. Sorry if I confused you, I was just goin off the top of me head.
i got this one, just did this today in math. so, what you first need to do is find the center of the hyperbolas. since there's a (Y-3) and an (X-2), the center is (2,3). ya dig? next, find the square roots of your denominators, in this case, 5 and 4. the square root of the first terms denominator is the distance from the center to the vertices, so therefore, your vertices are (2, 8) and (2, -2).

that's only for this problem though. if the X term had come first, it would have been (7,3) and (-3,3)
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Maths...

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square root of the denominators.
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The square root of the denominators will help determine the asymptotes, not the vertices's.

When i did hyperbolas, the vertex was always (0,0). According to this, the vertex is (2,3), which confuses me, because i never did that.

Now what does make sense is that you (or the book) made a mistake and meant to put a plus between the two fractions, creating an ellipse, not a hyperbola.

Call me crazy, but i think i am right.