Rewrite in standard form as:
(x/2)^2 - (y/3)^2 = 1
(i) Center is (0,0)
The vertices (lowest |x| value) are where y = 0, which would be (-2,0) amd (+2,0)
(ii) Foci are at
(0,-c) and (0,c), where
c = sqrt(2^2 + 3^2) = sqrt13 = 3.606
Eccentricity is defined as
e = c/a
In this case, a = 2, so e = c/2 = 1.803
(iii) Asymptotes are
y = (3/2)x and y = -(3/2)x
Give the equation of the hyperbola is 9x^2-4y^2=36.find
(i)center and vertices
(ii)foci and eccentricity
(iii)equation of assymptotes
1 answer
1 answer