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anonymous

  • 5 years ago

how do i get the center, vertices, foci, and eccentricity of 9x^2+4y^2-36x+8y_31=0?

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  1. amistre64
    • 5 years ago
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    it looks like the equation for a circle or ellipse

  2. amistre64
    • 5 years ago
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    (9x^2 -36x +n) + (4y^2 +8y +m) = 31 +n+m complete the squares to find n and m

  3. amistre64
    • 5 years ago
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    n = (36/2)^2 m=(8/2)^2 (9x^2 -36x +18^2) + (4y^2 +8y + 4^2) = 31+18^2+4^2

  4. amistre64
    • 5 years ago
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    9(x^2 -4x + 9^3) + 4(y^2 +2y + 4) = 371.... if i did it right in me head

  5. amistre64
    • 5 years ago
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    (3x - 18)^2 + (2y +4)^2 = 371 3(x-6)^2 + 2(y+2)^2 = 371 is what I ge so far

  6. amistre64
    • 5 years ago
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    divide evertyhing by 371 and see what you can work out I guess...

  7. amistre64
    • 5 years ago
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    and dont do stupid mistakes like I did..... that 3 and 2 dont get pulled out....

  8. amistre64
    • 5 years ago
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    9(x^2 -4x + 9^3) + 4(y^2 +2y + 4) = 371 is the correct way... solve your quadratics to get 2 solutions.....

  9. amistre64
    • 5 years ago
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    im gonna get it one way or another..... ignore that 9^3 flub... 9(x^2 -4x + 36) + 4(y^2 +2y + 4) = 371

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