Asked by Anonymous
write the equation of the hyperbola in graphing form from the given information:
Vertices at (2,0) and (-2,0); foci at (3,0) and (-3,0)
Vertices at (2,0) and (-2,0); foci at (3,0) and (-3,0)
Answers
Answered by
Damon
center at origin by symmetry
x^2/a^2 - y^2/b^2 = 1
transverse axis = 2 a = 2-(-2) = 4
so a = 2
now
x^2/4 -y^2/b^2 = 1 just find b from focus
center to focus = sqrt(a^2+b^2)
so 3 = sqrt(4+b^2)
9 = 4 + b^2
b = sqrt 5
so in the end
x^2/4 - y^2/5 = 1
x^2/a^2 - y^2/b^2 = 1
transverse axis = 2 a = 2-(-2) = 4
so a = 2
now
x^2/4 -y^2/b^2 = 1 just find b from focus
center to focus = sqrt(a^2+b^2)
so 3 = sqrt(4+b^2)
9 = 4 + b^2
b = sqrt 5
so in the end
x^2/4 - y^2/5 = 1
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