Question

Find the center, vertices, foci, and the equations of the asymptotes of the hyperbola. (If an answer does not exist, enter DNE.)

x2 − 2.25y2 + 22.5y − 92.25 = 0

Answer 1

Find the center, vertices, foci, and the equations of the asymptotes of the hyperbola. (If an answer does not exist, enter DNE.)

x^2 − 2.25y^2 + 22.5y − 92.25 = 0

x^2 − 2.25y^2 + 22.5y = 92.25

x^2 − 2.25(y^2 -10y) = 92.25

x^2 − 2.25(y^2 -10y+25) = 92.25 - (2.25 x 25)

x^2 − 2.25(y-5)^2 = 92.25 - 56.25

x^2 − 2.25(y-5)^2 = 36

Divide each term by 36,

(x^2/36) − (2.25/36)(y-5)^2 = 36/36

(x^2/36) − (y-5)^2 /16 = 1

is equivalent to

For a hyperbola with standard form of equation

It is a horizontal parabola with the following characteristics mainly,

Center : (h,k)

Length of transverse axis : 2a

Length of conjugate axis : 2b

Vertices : (h+a,k) (h-a,k)

Co-vertices : (h,k+b) (h,k-b)

a^2 + b^2 = c^2

Foci : (h+c,k) (h-c,k)

Equations of asymptotes : (y-k) = ±(b/a) (x-h)

a : 6, b : 4, c : sqrt 52 or 2sqrt13, h : 0, k : 5

Center : (0,5)

Vertices : (-6,5) (6,5)

Foci : (-2sqrt13,5) (2sqrt13,5)

Equations of asymptotes : (y-5) = ±(4/6)(x)

(y-5) = ±(2/3)(x)

y = (2/3) x + 5

y = (-2/3) x + 5