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

Question

(y-4)^2/16 - (x+2)^2/9 = 1?

Answer 1

The hyperbola equation is (y - 4)²/16 - (x + 2)²/9 = 1.

The standard form of the equation of a hyperbola with center (h, k) (where a and b are not equals to 0) is (x - h)²/a² - (y - k)²/b² = 1 (Transverse axis is horizontal) or (y - k)²/a² - (x - h)²/b² = 1 (Transverse axis is vertical).

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

Compare the equation (y - 4)²/16 - (x + 2)²/9 = 1 with (y - k)²/a² - (x - h)²/b² = 1.

a² = 16, b² = 9, k = 4 and h = - 2.

a = ± 4 and b = ± 3.

To find the value of c, substitute the value of a² = 16 and b² = 9 in b² = c² - a².

9 = c² - 16

25 = c²

c = ± 5.

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

The asymptote equations are y = ± 4/3 x.

The slopes of asymptote are ± 4/3,

Center = (h, k) = (- 2, 4),

Vertices = (h, k ± a) = (- 2, 4± 4) = (- 2, 8) and (- 2, 0)

Foci = (h, k ± c) = (- 2, 4 ± 5) = (- 2, 9) and (- 2, - 1).