Write the equation of the hyperbola whose vertices are at (0,±4) and whose foci are at (0,±11)

Question

amistre64 · Answer

trying for  asecond opinion; iwould too ;)

anonymous · Answer

$$x ^{2}/a -y ^{2}/b =1$$a=b=distance from vertices to center. Difference from center is 4. So your a and b should be 4.

anonymous · Answer

the transverse axis is vertical.

anonymous · Answer

The transverse axis is vertical. Thus the standard form of the hyperbola: 
y^2/a^2-x^2/b^2=1

anonymous · Answer

Oh yeah, your be is 4, since 4^2 is 16

anonymous · Answer

b

amistre64 · Answer

11^2 - 4^2 = b^2 right?

amistre64 · Answer

b^2 = 105 then

amistre64 · Answer

c = foci length from center right?

amistre64 · Answer

up 11 and down 11; which to me means the hypotenuse of the a-b triangle

amistre64 · Answer

the transverse in a hyper is got a smaller number under it as well

anonymous · Answer

hi amistre64 i dont mean to be a bother but I really needed your help for this one question!<3

amistre64 · Answer

the graphics one?

anonymous · Answer

yes!

amistre64 · Answer

ill try it...

anonymous · Answer

thank you!

anonymous · Answer

c = foci length from center right

amistre64 · Answer

and foci is 11 here; and transverse is 4^2 = 16; so b^2 has to be greater than or equal to 16 at least

anonymous · Answer

ah, i had a vision problem.. yeah it is 11.
i am sorry.

anonymous · Answer

$$c=11, a=4$$
$$b ^{2}=121-16$$
$$b ^{2}=105$$
So, the equation would be y^2/16 - x^2/(105) =1