divide by 64 to get
x^2/4 - y^2/16 = 1
Review your material on hyperbolas, and see whether you can see how to get the information shown here:
http://www.wolframalpha.com/input/?i=hyperbola+x%5E2%2F4+-+y%5E2%2F16+%3D+1
What are the vertices, foci, and asymptotes of the hyperbola with the equation 16x^2-4y^2=64?
What would the formulas be to find the answers?
3 answers
They want the answr, not an explanation. Give them the answer! ( *raises fist in the air*) lol
Answer from brainly:
The hyperbola: x^2/a^2 - y^2/b^2 = 1
foci: (+/-c,0) = (+/-sqrt(5),0) where c^2 = 1^2+2^2
and vertices (+/-a,0) and
asimptotes: y = +/-(b/a)x so,
16x^2-4y^2=64
x^2/1^2 - y^2/2^2 = 4
here, a = 1; b = 2 and c = sqrt[(a^2+b^2)] = sqrt[1 + 4] = sqrt(5)
So, 16x^2-4y^2=64 has
foci: foci: (+/-c,0) = (+/-sqrt(5),0) and
vertices: (+/-a,0) = (+/-1,0) and
asimptotes: y = +/-(b/a)x = +/-(2/1)x = +/-2x
now you don't have to use up your few questions on brainly!
The hyperbola: x^2/a^2 - y^2/b^2 = 1
foci: (+/-c,0) = (+/-sqrt(5),0) where c^2 = 1^2+2^2
and vertices (+/-a,0) and
asimptotes: y = +/-(b/a)x so,
16x^2-4y^2=64
x^2/1^2 - y^2/2^2 = 4
here, a = 1; b = 2 and c = sqrt[(a^2+b^2)] = sqrt[1 + 4] = sqrt(5)
So, 16x^2-4y^2=64 has
foci: foci: (+/-c,0) = (+/-sqrt(5),0) and
vertices: (+/-a,0) = (+/-1,0) and
asimptotes: y = +/-(b/a)x = +/-(2/1)x = +/-2x
now you don't have to use up your few questions on brainly!
1 answer