Question

Find the vertices and foci of the hyperbola with equation quantity x plus one squared divided by sixteen minus the quantity of y plus five squared divided by nine = 1.


A. Vertices: (-5, 3), (-5, -5); Foci: (-5, -6), (-5, 4)

B. Vertices: (2, -5), (-4, -5); Foci: (-4, -5), (2, -5)

C. Vertices: (3, -5), (-5, -5); Foci: (-6, -5), (4, -5)

D. Vertices: (-5, 2), (-5, -4); Foci: (-5, -4), (-5, 2)

Answer 1

Answer:

C

Step-by-step explanation:

This hyperbola is a horizontal hyperbola of the standard form:

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

Since our equation is

$\frac{(x+1)^2}{16}-\frac{(y+5)^2}{9}=1$,

a = 4 and b = 3.

The coordinates for the vertices are (±a, 0) and

the coordinates for the foci are (±c, 0).

We have a, but we need c.  To find c, we use Pythagorean's Theorem:

$c^2=4^2+3^2$ or

$c^2=16+9$ giving us that

c = 5.

But these a and c values have to be figured from the center of the hyperbola which is located at (-1, -5).  

For the vertices, then, we add the a value of 4 and -4 to the x value of the center, which is -1.  The -5 remains, since the vertices and the foci are on the same transcersal axis which is the line y = -5.

For the foci, then, we add the c value of 5 to -1, and again the -5 remains in the y position.

Vertices: (-1+4, -5)-->(3, -5) and (-1-4, -5)-->(-5, -5)

Foci: (-1+5, -5)-->(4, -5) and (-1-5, -5)-->(-6, -5)

Choice C