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

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1. sirm3d Group Title

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 Group Title

where is 4y^2 in the first equation?

3. sirm3d Group Title

oh, its 4y2. sorry.

4. PaulinePar Group Title

its ok hehe .. but where did you get 9 and 16 in the 2nd equation?

5. PaulinePar Group Title

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 Group Title

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 Group Title

Thank you so much :)) nakalimutan ko na po kc hehehe .. Yung another question ko po help naman po.

8. PaulinePar Group Title

Sir thank you so much po sa lahat ng response nyo. sa lahat ng help nyo :))

9. sirm3d Group Title

it is not a paraboloid of revolution as i said in my previous post. it is an elliptic paraboloid