What are the vertices of the hyperbola whose equation is [tex]\frac{(y+6)^2}{81}-\frac{(x+8)^2}{49}=1[/tex]?

A. [tex]$(-17,-6)$[/tex] and [tex]$(1,-6)$[/tex]
B. [tex]$(-15,-6)$[/tex] and [tex]$(-1,-6)$[/tex]
C. [tex]$(-8,-15)$[/tex] and [tex]$(-8,3)$[/tex]
D. [tex]$(-8,-13)$[/tex] and [tex]$(-8,1)$[/tex]



Answer :

To find the vertices of the hyperbola given by the equation

[tex]\[ \frac{(y+6)^2}{81} - \frac{(x+8)^2}{49} = 1, \][/tex]

we need to recognize that it is in the standard form for a hyperbola with a vertical transverse axis. The general equation of a hyperbola with a vertical transverse axis is

[tex]\[ \frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1, \][/tex]

where \((h, k)\) is the center of the hyperbola.

1. Identify the center \((h, k)\) of the hyperbola:
- The given equation is written as \(\left(y + 6\right)^2\) and \(\left(x + 8\right)^2\), so:
- \(h = -8\),
- \(k = -6\).

2. Determine the value of \(a\):
- The value of \(a^2\) is the denominator of the first term, which is 81.
- Therefore, \(a = \sqrt{81} = 9\).

3. Calculate the vertices:
- Since the transverse axis is vertical, the vertices are located \(a\) units above and below the center.
- Starting from the center \((-8, -6)\), we add and subtract the value of \(a\):

- Vertex 1: \((h, k + a) = (-8, -6 + 9) = (-8, 3)\).
- Vertex 2: \((h, k - a) = (-8, -6 - 9) = (-8, -15)\).

Therefore, the coordinates of the vertices are:

[tex]\[ \boxed{(-8, -15) \text{ and } (-8, 3)}. \][/tex]

Hence, the correct answer is:

[tex]\[ (-8, -15) \text{ and } (-8, 3). \][/tex]