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PaulinePar

  • one year ago

Can you guys identify the quadric surface using completing the squares of this equation. x^2+4y^2-6x+8y+4z=0? Thankyou :))

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  1. sirm3d
    • one year ago
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    x^2 -6x +y^2 + 8y = -4z x^2-6x+9 + y^2+8y+16=-4z+25 (x-3)^2 + (y+4)^2=-4(z-25/4) it is a paraboloid of revolution, with vertex at (3,-4,25/4)

  2. PaulinePar
    • one year ago
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    where is 4y^2 in the first equation?

  3. sirm3d
    • one year ago
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    oh, its 4y2. sorry.

  4. PaulinePar
    • one year ago
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    its ok hehe .. but where did you get 9 and 16 in the 2nd equation?

  5. PaulinePar
    • one year ago
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    sir how can i sketch the level curve z=k for the specified values of k. this is the equation z=x^2+y; k=-2, -1, 0, 1, 2

  6. sirm3d
    • one year ago
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    completing squares, my dear. \[x^2+4y^2-6x+8y+4z=0\\(x^2-6x+9)+4(y^2+2y+1)=-4z+9+4\\(x-3)^2+4(y+1)^2=-4(z-\tfrac{13}{4})\\\frac{(x-3)^2}{4}+\frac{(y+1)^2}{1}=\frac{z-\tfrac{13}{4}}{-1}\]

  7. PaulinePar
    • one year ago
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    Thank you so much :)) nakalimutan ko na po kc hehehe .. Yung another question ko po help naman po.

  8. PaulinePar
    • one year ago
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    Sir thank you so much po sa lahat ng response nyo. sa lahat ng help nyo :))

  9. sirm3d
    • one year ago
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    it is not a paraboloid of revolution as i said in my previous post. it is an elliptic paraboloid

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