12x^2-3y^2+24y-84=0 give the exact coordinates of any four points on the hyperbola

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12x^2-3y^2+24y-84=0 give the exact coordinates of any four points on the hyperbola

Mathematics
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hi can u help me
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
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

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Other answers:

ok can u help me with one more
(sqrt(3), 0) (-sqrt(3),0) are 2 that i get
that might nota worked out ....
yeah
i think i got lost trying to type it all in :)
ok
but can u help me
once you have \[\frac{x^2}{3}-\frac{(y-4)^2}{12}=1\] cant you just pick y and solve for x?
y =0
lets both try that together can u please show me
say for example \[y=2\] \[\frac{x^2}{3}-\frac{4}{12}=1\]
ok
\[\frac{x^2}{3}=1+\frac{4}{12}=1+\frac{1}{3}=\frac{4}{3}\]
i already gave you 4 points
\[x^2=4\] \[x=\pm 2\]
or you can pick any other y to solve for x.
so that is one point
actually 2: \[(-2,4)\] and \[(2,4)\]
(+-sqrt(7), 0) (+-sqrt(21)/2, 1) (+-2, 2) there are 6 points
let y = 1 \[\frac{x^2}{3}-\frac{(1-4)^2}{12}=1\] \[\frac{x^2}{3}-\frac{9}{12}=1\] \[\frac{x^2}{3}=1+\frac{9}{12}=1+\frac{3}{4}=\frac{7}{4}\]
so \[x^2=\frac{21}{4}\] \[x=\pm \frac{\sqrt{21}}{2}\]
got it?

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