Question

12x^2-3y^2+24y-84=0

give the exact coordinates of any four points on the hyperbola

Answer 1

again?

Answer 2

yes ou didnt give me a good answer

Answer 3

it wasnt right

Answer 4

hi can u help me

Answer 5

lets see what we can accomplish

Answer 6

12x^2-3(y^2-8y)=84
12x^2-3(y-4)^2=100

Answer 7

no it was wrong... forget it :)

Answer 8

(12x^2) + (-3y^2+24y) -84=0

(12x^2) +-3 (y^2-8y + ___) =84 +-3(____)

(12x^2) +-3 (y^2-8y + 16) =84 +-3(16)

12x^2 -3 (y-4)^2 =84 -48 = 36

12x^2 3(y-4)^2
----- - -------- = 1
36 36

(1/3)x^2 - (1/12)(y-4^2) = 1

Answer 9

ok

Answer 10

when y = 4; x = +- sqrt(3) so..

(+sqrt(3),4) (-sqrt(3),4) are 2 points we know of; they are the vertices of the hyperola right?

Answer 11

ok

Answer 12

when x=3; we get 3-y = 1

y will have to even out to 2 then...

(y-4)^2
------ = 2
12

(y-4)^2 = 24

y^2 -8y +16-24 = 0

y^2 -8y -8 = 0

y = -4 +- 2sqrt(2) can you input a sum of numbers for a point?

Answer 13

i dk

Answer 14

i might be over thinking that; lets try at y = 0

(1/3)x^2 - 16/12 = 1

x^2 = (12+16/4)

x^2 = 7

x = +- sqrt(7) then is simpler

(sqrt(7),0) (-sqrt(7),0) are another 2 points

Answer 15

hi. didn't we get exactly this answer before? was it wrong? because it looks good to me.

Answer 16

sqrt(3) need to dbl chk it

Answer 17

36 -48+96 - 84 ?=0

Answer 18

132 -132 = 0 is good

Answer 19

sqrt(7) is good cause 12.7 =84

Answer 20

good thanks

Answer 21

can u help me with one more

Answer 22

4x^2-6y^2+32x-24y+16=0

Answer 23

give the exact coordinated of an four points

Answer 24

can u helpme