Question

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

give the exact coordinates of any four points on the hyperbola

Answer 1

hi can u help me

Answer 2

if x=0
-3y^2 +24y -84 =0
factor out a -3
y^2 -8y + 28 =0
b^2 -4ac < 0, complex solutions

if y=0
12x^2 = 84
x^2 = 7
x = +-sqrt(7)
that gives 2 points

get x by itself
12x^2 = 3y^2 -24y +84
x^2 = 1/4y^2 - 2y +7
x = sqrt[1/4y^2 -2y +7]
pick 2 y_values
y=1 --> x =+- sqrt(21/4) = +-sqrt(21)/2
there are other 2 points

Answer 3

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

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

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

Answer 4

ok can u help me with one more

Answer 5

(sqrt(3), 0) (-sqrt(3),0) are 2 that i get

Answer 6

that might nota worked out ....

Answer 7

yeah

Answer 8

i think i got lost trying to type it all in :)

Answer 9

ok

Answer 10

but can u help me

Answer 11

once you have \[\frac{x^2}{3}-\frac{(y-4)^2}{12}=1\] cant you just pick y and solve for x?

Answer 12

y =0

Answer 13

lets both try that together can u please show me

Answer 14

say for example
\[y=2\]
\[\frac{x^2}{3}-\frac{4}{12}=1\]

Answer 15

ok

Answer 16

\[\frac{x^2}{3}=1+\frac{4}{12}=1+\frac{1}{3}=\frac{4}{3}\]

Answer 17

i already gave you 4 points

Answer 18

\[x^2=4\]
\[x=\pm 2\]

Answer 19

or you can pick any other y to solve for x.

Answer 20

so that is one point

Answer 21

actually 2:
\[(-2,4)\] and \[(2,4)\]

Answer 22

(+-sqrt(7), 0)
(+-sqrt(21)/2, 1)
(+-2, 2)
there are 6 points

Answer 23

let y = 1
\[\frac{x^2}{3}-\frac{(1-4)^2}{12}=1\]
\[\frac{x^2}{3}-\frac{9}{12}=1\]
\[\frac{x^2}{3}=1+\frac{9}{12}=1+\frac{3}{4}=\frac{7}{4}\]

Answer 24

so
\[x^2=\frac{21}{4}\]
\[x=\pm \frac{\sqrt{21}}{2}\]

Answer 25

got it?