## PaulinePar 2 years 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

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

where is 4y^2 in the first equation?

3. sirm3d

oh, its 4y2. sorry.

4. PaulinePar

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

5. PaulinePar

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

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

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

8. PaulinePar

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

9. sirm3d

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