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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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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)
where is 4y^2 in the first equation?
oh, its 4y2. sorry.

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its ok hehe .. but where did you get 9 and 16 in the 2nd equation?
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
completing squares, my dear. \[x^2+4y^2-6x+8y+4z=0\\(x^2-6x+9)+4(y^2+2y+1)=-4z+9+4\\(x-3)^2+4(y+1)^2=-4(z-\tfrac{13}{4})\\\frac{(x-3)^2}{4}+\frac{(y+1)^2}{1}=\frac{z-\tfrac{13}{4}}{-1}\]
Thank you so much :)) nakalimutan ko na po kc hehehe .. Yung another question ko po help naman po.
Sir thank you so much po sa lahat ng response nyo. sa lahat ng help nyo :))
it is not a paraboloid of revolution as i said in my previous post. it is an elliptic paraboloid

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