SOLUTION: Find the vertices, foci, and the equations of the asymptotes of the hyperbola (x^2/16)+(y^2/64)=1

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Question 972995: Find the vertices, foci, and the equations of the asymptotes of the hyperbola (x^2/16)+(y^2/64)=1

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2%2F16%2By%5E2%2F64=1 ....this is an ellipse, not hyperbola
looking at the denominators, I see that a%5E2+=+16 and b%5E2+=+64,
so
semi-minor axis length a+=+4
and
semi-major axis length b+=+8
the equation c%5E2+=a%5E2+%2Bb%5E2 tells me that
c%5E2+=+16+%2B+64+=+80,
so c+=sqrt%28+80%29=sqrt%28+4%2A4%2A5%29=2%2A2sqrt%28+5%29
c=4sqrt%28+5%29

since %28x-0%29%5E2+=x%5E2 and %28y-0%29%5E2=y%5E2+, then the center is at
(h, k) = (0, 0)

the vertices are at (0, -a) and (0,+a)
=> (0,+-4) and (0,+4),
and the foci at (0,sqrt%28b%5E2-a%5E2%29)
sqrt%28b%5E2-a%5E2%29
=sqrt%2864-16%29
=sqrt%2848%29
=sqrt%284%5E2%2A3%29
=4sqrt%283%29
(0, -4sqrt%283%29) and (0, 4sqrt%283%29)

asymptotes:
Ellipses do not have asymptotes. Hyperbolas do, but Ellipses do not.




if it’s minus
x%5E2%2F16-y%5E2%2F64=1 then you have hyperbola
and a=4, b=8, c=4sqrt%285%29
so,
foci is at:
(-4sqrt%285%29, 0) and (4sqrt%285%29, 0)
or (-8.9, 0) and (8.9, 0)
vertices: (-4,+0) and (4, 0)
center :(0, 0)
semimajor axis length: +4
semiminor axis length: 8
eccentricity :c%2Fa=>4sqrt%285%29%2F4=+sqrt%285%29 or 2.2
asymptotes :y+=+2+x and y+=+-2+x