Question

Find the vertices and foci of the hyperbola with equation x squared over nine minus y squared over sixteen = 1.

Answer 1

Explanation:

equation x squared over nine minus y squared over sixteen = 1 is the same as

x^2/9 - y^2/16 = 1

x is at its min/max when y^2=0, y=0

x^2/9 = 1

x = +3

vertices (+3, 0)

c^2 = 9 + 16 = 25

c = +5

foci (+5, 0)

Answer 2

The vertices are at (3, 0) and (-3, 0).

The foci are at (5, 0) and (-5, 0).

Explanation:

Compare given equation x^2/9 - y^2/16 = 1

against standard hyperbola form (x - h)^2 / a^2 - (y - k)^2 / b^2 = 1,

the hyperbola opens left and right with its center (h, k) at (0, 0).

a^2 = 9 so a = +/-3

The vertices are at (3, 0) and (-3, 0).

a^2 + b^2 = c^2 where c is 2x distance between foci.

9 + 16 = c^2

c = +/-5

The foci are at (5, 0) and (-5, 0).