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Mathematics 6 Online
OpenStudy (anonymous):

Will Medal, just really need help!! Find an equation in standard form for the hyperbola with vertices at (0, ±4) and asymptotes at y = ±1 divided by 3.x.

OpenStudy (anonymous):

looks like you are stuck with analytic geometry :) now lets start with how standard form looks like\[\frac{ y^2 }{ a^2 } - \frac{ x^2 }{ b^2 } = 1\]

OpenStudy (anonymous):

Ok @saseal

OpenStudy (anonymous):

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OpenStudy (anonymous):

My answer choices are: A. y squared over 144 minus x squared over 16 = 1 B. y squared over 16 minus x squared over 36 = 1 C. y squared over 16 minus x squared over 144 = 1 D. y squared over 36 minus x squared over 4 = 1 I'm between B and C

OpenStudy (anonymous):

now we have the gradient y=a/b which is 1/3

OpenStudy (anonymous):

and we know a is 4 so b gotta be 12 since the gradient have to match up

OpenStudy (anonymous):

you can guess which one it is by now :)

OpenStudy (anonymous):

C? @saseal

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

Can you help me with a few others?

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

Thanks! @saseal Find the center, vertices, and foci of the ellipse with equation x squared divided by 400 plus y squared divided by 625 = 1.

OpenStudy (anonymous):

\[\frac{ x^2 }{ 400 } + \frac{ y^2 }{ 625 } = 1\] this should be easy by now

OpenStudy (anonymous):

\[\frac{ x^2 }{ 20^2 } + \frac{ y^2 }{ 25^2 } = 1\]

OpenStudy (anonymous):

where do you think is the major axis of this egg?

OpenStudy (anonymous):

major axis is the longer side

OpenStudy (anonymous):

The y-axis

OpenStudy (anonymous):

correct

OpenStudy (anonymous):

now think where's the center

OpenStudy (anonymous):

theres no h and k here

OpenStudy (anonymous):

0,0

OpenStudy (anonymous):

there we have the both the center and vertices now

OpenStudy (anonymous):

What's the vertices though?

OpenStudy (anonymous):

center (0, 0) vertices (0, ±25)

OpenStudy (anonymous):

Oh ok!

OpenStudy (anonymous):

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OpenStudy (anonymous):

now we need to find c to get the foci

OpenStudy (anonymous):

\[c=\sqrt{a^2-b^2}\]

OpenStudy (anonymous):

Ok, so C = 15? @saseal

OpenStudy (anonymous):

+/- 15

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

Ok so would my answer be: Center: (0, 0); Vertices: (-25, 0), (25, 0); Foci: (-15, 0), (15, 0)

OpenStudy (anonymous):

Or no. It would be: Center: (0, 0); Vertices: (0, -25), (0, 25); Foci: (0, -15), (0, 15

OpenStudy (anonymous):

Could you help me with one more? @saseal

OpenStudy (anonymous):

yea

OpenStudy (anonymous):

sry im not looking at the pc all the time

OpenStudy (anonymous):

No problem. And I'm in a new chapter so: Find the derivative of f(x) = 6 divided by x at x = -2.

OpenStudy (anonymous):

And also: Find the derivative of f(x) = 4x + 7 at x = 5. @saseal

OpenStudy (anonymous):

sry i fell asleep

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