Question

What are the vertices of the hyperbola whose equation is (x+3)^2/4 - (y-5)^2/16 =1

Answer 1

Answer:B

Step-by-step explanation:

Answer 2

The vertices of the hyperbola are (−1,5) and (−5,5)

What is hyperbola?

Hyperbola is defined as a conic section, two-branched open curve, produced by the intersection of a circular cone.

Here the equation of hyperbola is

$\dfrac{(x+3)^2}{4} - \dfrac{(y-5)^2}{16} =1$

If the general equation of hyperbola is

$\dfrac{(x-h)^2}{a^2} - \dfrac{(y-k)^2}{b^2} =1$

$\dfrac{(x+3)^2}{2^2} - \dfrac{(y-5)^2}{4^2} =1$

Here, a=2, b=4,  and h=−3, k=5

Vertices are ( h±a, k)

Vertices are (−3±2,5) i.e. (−1,5) and (−5,5)

Learn more about hyperbola here:

https://brainly.com/question/27799190

#SPJ2