Question

What is the vertices, foci, and slope of the asymptotes for the hyperbola whose equation is, x^2/81 - y^2/36 = 1?

Answer 1

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).