Question

Which features describe the graph of y^2/96^2 - x^2/40^2 =1? Check all that apply.

a focus at (104, 0)
a focus at (0,−96)
a vertex at (−40, 0)
a vertex at (0, 96)
the center at (0, 0)

Answer 1

its the last two

a vertex at (0, 96)

the center at (0, 0)

Explanation:

just took the test.

Answer 2

The given equation is

`(y^(2))/(96^(2)) - (x^(2))/(40^(2)) =1`

From the given equation,
a = 40
b = 96
(h,k) = (0,0), the center

The vertices are at (0,96) and at (0, -96).
The curves open upward and downward.

c² = a² + b² = 40² + 96² = 10816
c = 104
Therefore the foci are at (0, -104) and (0, 104).

Check the given answers:
1. a focus at (104, 0) is INCORRECT
2. a focus at (0, -96) is INCORRECT
3. a vertex at (-40, 0) is INCORRECT
4. a vertex at (0, 96) is CORRECT
5. the center at (0, 0) is CORRECT