Question

find the centre, eccentricity , vertices and foci of the hyperbola 12x^2-4y^2-24x+32y-127=0

Answer 1

Answer:

Step-by-step explanation:

12(x² - 2x + 1) - 12 - 4(y² - 8y + 16) + 64 - 127 = 0

12(x - 1)² - 4(y - 4)² = 75

$\frac{4(x-1)^2}{25}$ - $\frac{4(y-4)^2}{75}$ = 1

The center (h, k) is at ( 1, 4 )

a² = $\frac{25}{4}$ and b² = $\frac{75}{4}$  

a = ± $\frac{5}{2}$ and b = ± $\frac{5\sqrt{3} }{2}$  

c² = a² + b²

c² = 25 ; c = ± 5

The formula of an eccentricity is

e = $\frac{\sqrt{a^2 +b^2} }{a}$ = 5 ÷ $\frac{5}{2}$ = 2  

The coordinates of the vertices are (h ± a, k)

( 1 + $\frac{5}{2}$ , 4 ) = (3.5, 4)

( 1 - $\frac{5}{2}$ , 4) = ( - 1.5 , 4)

The coordinates of the foci are (h ± c , k)

( 1 - 5 , 4) = ( - 4, 4)

(1 + 5 , 4) = ( 6, 4 )