Answer:
The center is (8 , -9)
The vertices are (11 , -9) and (5 , -9)
The foci are (8 , -9 + √58) and (8 , -9 - √58)
The equations of the asymptotes are y = 3/7(x − 8) - 9 , y = -3/7 (x − 8) - 9
Step-by-step explanation:
- The standard form of the equation of a hyperbola with
center (h , k) and transverse axis parallel to the y-axis is
(y - k)²/a² - (x - h)²/b² = 1
- The length of the transverse axis is 2 a
- The coordinates of the vertices are ( h ± a , k )
- The length of the conjugate axis is 2 b
- The coordinates of the co-vertices are ( h , k ± b )
- The coordinates of the foci are (h , k ± c), where c² = a² + b²
- The equations of the asymptotes are y = ± a/b (x − h) + k
* Now lets solve the problem
∵ (y + 9)²/9 - (x - 8)²/49 = 1
∴ h = 8 and k = -9
∴ a² = 9 ⇒ a = ± 3
∴ b² = 49 ⇒ b = ± 7
∵ c² = a² + b²
∴ c² = 9 + 49 = 58
∴ c = ± √58
∵ The center is (h , k)
∴ The center is (8 , -9)
∵ The coordinates of the vertices are ( h ± a , k )
∴ The vertices are (8 + 3 , -9) and (8 - 3 , -9)
∴ The vertices are (11 , -9) and (5 , -9)
∵ The coordinates of the foci are (h , k ± c)
∴ The foci are (8 , -9 + √58) and (8 , -9 - √58)
∵ The equations of the asymptotes are y = ± a/b (x − h) + k
∴ The equations of the asymptotes are y = 3/7 (x - 8) - 9 and
y = -3/7 (x − 8) - 9
