Question

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

Answer 1

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)

Answer 2

where is 4y^2 in the first equation?

Answer 3

oh, its 4y2. sorry.

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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