Respuesta :
Answer:
- Vertices:(-14,4) and (10,4).
- Foci: (–17, 4) and (13, 4)
Step-by-step explanation:
Given the equation of the hyperbola
[tex]\dfrac{(x+2)^2}{144}-\dfrac{(y-4)^2}{81} =1[/tex]
Since the x part is added, then
[tex]a^2=144; b^2=81\\a=12,b=9[/tex]
Also, this hyperbola's foci and vertices are to the left and right of the center, on a horizontal line paralleling the x-axis.
From the equation, clearly the center is at (h, k) = (–2, 4). Since the vertices are a = 12 units to either side, then they are at (-14,4) and (10,4).
From the equation
[tex]c^2=a^2+b^2=144+81=225\\c=15[/tex]
The foci, being 15 units to either side of the center, must be at (–17, 4) and (13, 4)
