toxicsugar22
  • toxicsugar22
12x^2-3y^2+24y-84=0 give the exact coordinates of any four points on the hyperbola
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
again?
toxicsugar22
  • toxicsugar22
yes ou didnt give me a good answer
toxicsugar22
  • toxicsugar22
it wasnt right

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More answers

toxicsugar22
  • toxicsugar22
hi can u help me
amistre64
  • amistre64
lets see what we can accomplish
watchmath
  • watchmath
12x^2-3(y^2-8y)=84 12x^2-3(y-4)^2=100
watchmath
  • watchmath
no it was wrong... forget it :)
amistre64
  • amistre64
(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
toxicsugar22
  • toxicsugar22
ok
amistre64
  • amistre64
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?
toxicsugar22
  • toxicsugar22
ok
amistre64
  • amistre64
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?
toxicsugar22
  • toxicsugar22
i dk
amistre64
  • amistre64
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
anonymous
  • anonymous
hi. didn't we get exactly this answer before? was it wrong? because it looks good to me.
amistre64
  • amistre64
sqrt(3) need to dbl chk it
amistre64
  • amistre64
36 -48+96 - 84 ?=0
amistre64
  • amistre64
132 -132 = 0 is good
amistre64
  • amistre64
sqrt(7) is good cause 12.7 =84
toxicsugar22
  • toxicsugar22
good thanks
toxicsugar22
  • toxicsugar22
can u help me with one more
toxicsugar22
  • toxicsugar22
4x^2-6y^2+32x-24y+16=0
toxicsugar22
  • toxicsugar22
give the exact coordinated of an four points
toxicsugar22
  • toxicsugar22
can u helpme

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