Find an equation for the hyperbola.

[wat2000]

Find an equation for the hyperbola that satisfies the given conditions. Foci (0, ±7), length of transverse axis 7. Im a little confused on how to solve this. I tried to solve it and i've found that c^2= 49 so I know that a^2 and b^2 must add up to 49 but im not sure what my next step is. Can someone help me?

[LCKurtz]

Find an equation for the hyperbola that satisfies the given conditions. Foci (0, ±7), length of transverse axis 7. Im a little confused on how to solve this. I tried to solve it and i've found that c^2= 49 so I know that a^2 and b^2 must add up to 49 but im not sure what my next step is. Can someone help me?

If by "the length of the transverse axis" you mean the distance between the intercepts, then you would have 2a = 7. That should help.

[wat2000]

so a would equal 2/7? Do i solve 2a=7 to create that fraction?

[eumyang]

so a would equal 2/7? Do i solve 2a=7 to create that fraction?

You mean 7/2. :cry: Now find b.

[js14]

yes I mean 7/2. anyway i dont know how to find b. the only thing i can think of is to square 7/2 and 7 to give me 7/4 and 49 and then subtract them to give me b. that doesnt work though. whats my next step?

[eumyang]

yes I mean 7/2. anyway i dont know how to find b. the only thing i can think of is to square 7/2 and 7 to give me 7/4 and 49 and then subtract them to give me b. that doesnt work though. whats my next step?
Yes, you need to use the Pythagorean relation
c^2 = a^2 + b^2
to find b. What do you mean, it "doesn't work"? Show us what you got.

[js14]

ok I have it now. i was looking at it wrong. 7/2 squares to 49/4 so a^2=49/4. and when i use the formula c^2 = a^2 + b^2 to find b^2. i plug 7 into c^2 and 49/4 into a^2. this gives me (7)^2=49/4 +b^2. then i square 7 to give me 49=49/4+b^2 then subtract 49/4 to get b^2 by itself and I get 147/4 = b^2. and since the foci points are (0, ±7) the y^2 should be out front.
y^2/(49/4) + x^2/(147/4) =1. and this simplifies to 4y^2/49 + 4x^2/147 = 1. and thats the final answer.