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What is the vertices, foci, and slope of the asymptotes for the hyperbola whose equation is,

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What is the vertices, foci, and slope of the asymptotes for the hyperbola whose equation is, x^2/81 - y^2/36 = 1?
asked Mar 17, 2014 in ALGEBRA 2 by homeworkhelp Mentor

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The hyperbola equation is x2/81 - y2/36 = 1.

The standard form of the equation of a hyperbola with center at the origin (where a and b are not equals to 0) is x2/a2 - y2/b2 = 1 (Transverse axis is horizontal) or y2/a2 - x2/b2 = 1 (Transverse axis is vertical).

The vertices and foci are, respectively a and c units from the center and the relation between a, b and c is b2 = c2 - a2.

Compare the equation x2/81 - y2/36 = 1 with x2/a2 - y2/b2 = 1.

a2 = 81 and b2 = 36.

a = ± 9 and b = ± 6.

To find the value of c, substitute the value of a2 = 81 and b2 = 36 in b2 = c2 - a2.

36 = c2 - 81

117 = c2

c = ± √117.

Here the transverse axis is horizontal, the asymptotes are of the forms y = (b/a) x and y = - (b/a) x.

The asymptotes equations are y = ± 6/9 x.

The slope asymptote is ± 6/9 = ± 2/3, Vertices = (± a, 0) = (± 9, 0) and Foci = (± c, 0) = (± √117, 0).

 

answered Mar 27, 2014 by steve Scholar

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