#1
hi. i forgot to find the verticies of the hyperbolas. i really feel stupid
(Y-3)^2 /25 - (X-2)^2 /16
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#3
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
#4
^ what he said
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#5
Why 8?
im my book its givin me 2,8 and 2,-2
(\__/) I became insane, with long intervals of horrible
(='.'=) sanity
(")_(")
#6
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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#7
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.
#8
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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#10
square root of the denominators.
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#13
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.