An equation of a hyperbola is given 16X2 32x 4y2 16y 64

An equation of a hyperbola is given. 16X^2 + 32x - 4y^2 + 16y + 64 = 0 (a) Find the center, vertices, foci, and asymptotes of the hyperbola. (Enter your asymptotes as a comma-

Solution

Equation of the hyperbola: 16x2+32x4y2+16y+64=016x2+32x4y2+16y+64=0 or 4x2+8xy2+4y+16=04x2+8xy2+4y+16=0.

Standard form: 14(x+1)2+116(y2)2=114(x+1)2+116(y2)2=1.

Center: (1,2)(1,2).

Vertices: (1,2)(1,2), (1,6)(1,6).

Co-vertices: (3,2)(3,2), (1,2)(1,2).

Foci: (1,25+2)(1,2.47213595499958)(1,25+2)(1,2.47213595499958), (1,2+25)(1,6.47213595499958)(1,2+25)(1,6.47213595499958).

First asymptote: y=2x=2.0xy=2x=2.0x

Second asymptote: y=2x+4=2.0x+4.0