Question

Find an equation in standard form for the hyperbola with vertices at (0, ±6) and foci at (0, ±9). ??? I think its (y^2/36)-(x^2/81)=1?

Answer 1

not quite, the "y" part is correct
for hyperbola:
\[a^{2} +b^{2} = c^{2}\]
where "c" is foci distance
\[36+b^{2} = 81\]
\[b^{2} = 45\]

Answer 2

ohhh okay,wow lol thankyou(:

Answer 3

yw