Question

Algebra 2b ??

Answer 1

@jim_thompson5910

Answer 2

The center is (0,0) since this the midpoint of the two points (-5,0),(5,0)

Answer 3

okaay :)

Answer 4

The distance from (0,0) to (5,0) is 5 units. So this is the value of 'a'

a = 5

Answer 5

i have a few more after this one!

Answer 6

The focal distance is 6 units because (6,0) is 6 units away from (0,0)

therefore c = 6

Answer 7

c^2 = a^2 + b^2

6^2 = 5^2 + b^2

36 = 25 + b^2

36 - 25 = b^2

11 = b^2

b^2 = 11

Answer 8

So a^2 = 25, b^2 = 11

Answer 9

The hyperbola opens left/right, which means

(x^2)/(a^2) - (y^2)/(b^2) = 1

(x^2)/(25) - (y^2)/(11) = 1

is the equation of this hyperbola

Answer 10

okay next one?

Answer 11

@jim_thompson5910

Answer 12

sure I'll help with 2 more

Answer 13

oaky :(there it is

Answer 14

16x^2 - 4y^2 = 64

16x^2/64 - 4y^2/64 = 64/64 ... divide every term by 64

(x^2)/(4) - (y^2)/(16) = 1

The last equation is in the form (x^2)/(a^2) - (y^2)/(b^2) = 1 where

a^2 = 4 ---> a = 2

b^2 = 16 ---> b = 4

Answer 15

so

c^2 = a^2 + b^2

c^2 = 2^2 + 4^2

c^2 = 4 + 16

c^2 = 20

c = sqrt(20) <--- this is the focal distance

Answer 16

So using this info, can you tell me what the vertices and foci are?

Answer 17

Are you able to find them?

Answer 18

Or no?