What is the FOCI for this Hyperbola? Help check my work please?

0 ⤊

y^2/25-x^2/36=1

I get +- 6, 0 and sqrt61? Is this correct. Vertices I get are (0,5)(0,-5) So I use these for H and K? right? Thanks

2007-05-04 14:35:03 · 2 answers · asked by lawrence c 2 in Science & Mathematics ➔ Mathematics

Thanks for a prompt response

2007-05-04 14:50:16 · update #1

The center is (0,0), vertices (0, 5) and (0,-5). Foci are (0, ±√61)

2007-05-04 14:49:15 · answer #1 · answered by Kathleen K 7 · 0⤊ 0⤋

The focal axis is the y axis whose equation is x=0
c^2=a^2+b^2
c^2=61 so c=sqrt(61)
the foci are (0,sqrt61) and (0,-sqrt61)
The vertices are right

2007-05-04 21:43:48 · answer #2 · answered by santmann2002 7 · 0⤊ 0⤋