x^2-4y^2-6x-32y-64=0

Question

need vertices foci and asymptotes.

Answer 1

Given equation x^2-4y^2-6x-32y-64 = 0

x^2-6x-4y^2-32y-64 = 0

To complete squre add half of x coefficient^2 to each side.

Add 9 to each side.

x^2-6x-4y^2-32y-64+9 = 9

x^2-6x+9-4y^2-32y-64 = 9

(x-3)^2-4(y^2+8y+16) = 9

(x-3)^2-4(y+4)^2 = 9

Divide to each side by 9.

(x-3)^2/9 - 4(y+4)^2/9 = 1

(x-3)^2/9 - (y+4)^2/(9/4) = 1

(x-3)^2/3^2 - (y-(-4))^2/(3/2)^2 = 1

Horizontal hyperabola equation is (x-h)^2/a^2 - (y-k)^2/b^2 = 1

a = 3, b = 3/2

Center (h,k) = (3,-4)

Vertices (h+a,k) = (6,-4)

(h-a,k) = (0,-4)

c = √(a^2+b^2) = √(9+9/4) = √(45/4) = 3.35

Foci (h+c,k) = (6.35,-4)

(h-c,k) = (-0.35,-4)

Equation for the asymptotes y = ±b/a(x-h)+k

y = ±(3/2)/3[x-3]-4

y = (1/2)(x-3)-4 and y = -(1/2)(x-3)-4

y = x/2-3/2-4 and y = -x/2+3/2-4

y = x/2-11/2 and y = -x/2-5/2.