Question

more parabolas >:D

Answer 1

Find the vertices and foci of the hyperbola with equation quantity x minus three squared divided by sixteen minus the quantity of y plus four squared divided by nine = 1
@ganeshie8

Answer 2

\[\frac{ (X-3)^2 }{ 16 } - \frac{ (Y+4)^2 }{ 9 }\]

Answer 3

hyperbola with center away from (0, 0)

Answer 4

indeed c:

Answer 5

use the chart

Answer 6

so its (3+-4,2) for the verticies

Answer 7

and (3+- 5,2) for the foci right?

Answer 8

your hyperbola :

\(\large \dfrac{(x-3)^2}{16} - \dfrac{(y+4)^2}{9} = 1\)


which is same as :

\(\large \dfrac{(x-3)^2}{4^2} - \dfrac{(y-(-4))^2}{3^2} = 1\)

Answer 9

\(\implies\)

\(h = 3, k = -4\)
\(a = 4, b = 3\)

Answer 10

right ?

Answer 11

that gives :

Vertices = \((h\pm a, k) = (3 \pm 4, -4)\)

Answer 12

right ?

Answer 13

k is +4 because that negatives part of the equation so you dont count it.... but if it was written as +4... oh i see where i went wrong .-.

Answer 14

so verticies are (3+- (-4),-2)
and foci are (3+-5,-2

Answer 15

hey why are you putting k = -2 ?

Answer 16

k = -4 right ?

Answer 17

oh so its (3+-4,-4)
and (3+-5,-4)?

Answer 18

Yes !

Answer 19

which would be (1,-4) and (7,-4)
(-2,-4) and (8,-4)

Answer 20

*-1,-4 for the first one

Answer 21

which would be (\(\color{red}{-}\)1,-4) and (7,-4)
(-2,-4) and (8,-4)

Answer 22

:) looks good !

Answer 23

yay c: ive got one more quesiton i got that ill tag you in c: