Algebra question

Question

The hyperbola is given by equation :

4x² – 9y²+32x + 54y – 53 = 0.

Find the coordinates of the center, the foci, the eccentricity and the asymptoteso of the hyperbola.

Answer 1

4x² – 9y² + 32x + 54y – 53 = 0.

We complete the square and get 4x² + 32x + 64 - 64 – (9y^{2 -} 54y + 81 - 81) – 53 = 0

4(x+4)^2 - 9(y- 3)^2 = 36

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

The center is (-4, 3)

The foci are a distance √(9+4) = √13 from the center

They are at (-4 - √13, 3) (-4 + √13, 3)

As the slope of the asymptotes will be ±2/3, the equations of the asymptotes are

(y- 3) = ±2/3(x + 4) or

y = 2/3 x + 5 2/3 or 2/3x + 17/3

y = -2/3x + 1/3

e = √(a^2+b^2)/a = √13/3