Respuesta :
Answer:
1) The vertices of the hyperbola are (-8, -15), (-8, 3)
2) The foci of the hyperbola are (11, -7) or (1, -7)
Step-by-step explanation:
1) The vertices of a hyperbola given in the form [tex]\dfrac{\left (y - k\right)^2}{a^2} -\dfrac{\left (x - h\right)^2}{b^2} = 1[/tex] is (h, k ± a)
Therefore, for the given hyperbola with the following equation;
[tex]\dfrac{\left (y + 6\right)^2}{81} -\dfrac{\left (x + 8\right)^2}{49} = 1[/tex], we have;
[tex]\dfrac{\left (y + 6\right)^2}{9^2} -\dfrac{\left (x + 8\right)^2}{7^2} = 1[/tex]
Therefore the vertices are;
(-8, -6 ± 9) = (-8, -15), (-8, 3)
2) The equation of the given hyperbola is presented as follows;
[tex]\dfrac{\left (x - 6\right)^2}{16} -\dfrac{\left (y + 7\right)^2}{9} = 1[/tex]
The given hyperbola is of the form;
[tex]\dfrac{\left (x - h\right)^2}{a^2} -\dfrac{\left (y - k\right)^2}{b^2} = 1[/tex]
We get;
[tex]\dfrac{\left (x - 6\right)^2}{4^2} -\dfrac{\left (y + 7\right)^2}{3^2} = 1[/tex]
The foci is then (h ± c, k)
Therefore, by comparison, the foci of the given hyperbola is (6 ± c, -7)
Where c² = a² + b²
∴ c² = 4² + 3² = 25
c = √25 = 5
∴ The foci; (6 ± 5, -7) = (11, -7) or (1, -7).
