Question

Find an equation in standard form for the hyperbola with vertices at (0, +/-6) and asymptotes at
y = +/- 3/4 x

Answer 1

@iambatman @Compassionate @jigglypuff314 anyone????

Answer 2

http://www.analyzemath.com/HyperbolaProblems/HyperbolaProblems.html

Answer 3

um thanks, but I did read my book, I just don't really get it...

Answer 4

I don't know, sorry
perhaps @ganeshie8 or @iambatman can help you

Answer 5

thanks! :)

Answer 6

see if this helps

Answer 7

I see the equation I have to use... but how to I get a and b ?

Answer 8

@Turner from the data or your texr, I see that we have:
\[\frac{ b }{ a }=\frac{ 3 }{ 4 }\]
and
\[c=6\]
Now, in the case of an hyperbola, the relationship between a, b and c, is:
\[c ^{2}=a ^{2}+b ^{2}\]
From the first equation, we can write:
\[b=\frac{ 3 }{ 4 }a\]
please insert that relationship for b, into the quadratic equation forc, and you will get a and b.
the answer is: \[a=\frac{ 24 }{ 5 }\] and \[b=\frac{ 18 }{ 5 }\]
please, check all my statements

Answer 9

@Turner the standard form, as you know, is:
\[\frac{ x ^{2} }{ a ^{2} }-\frac{ y ^{2} }{ b ^{2} }=1\]

Answer 10

oh I see! So then I would write it like (x^2 / 4.8) - (y^2 / 3.6) ?

Answer 11

better is:
\[\frac{ 25x ^{2} }{ 576 }-\frac{ 25y ^{2} }{ 324 }=1\]