For the quadratic surface defined by 36x^2-72x-9y^2-36y-16z^2+96z-288=0
Put the surface into standard form and identify the center of the surface
complete the square ...
36(x^2 - 2x + ...) - 9(y^2 + 4y + ....) - 16(z^2 - 6z + ...) = 288
36(x^2 - 2x + 1) - 9(y^2 + 4y + 4) - 16(z^2 - 6z + 9) = 288 + 36 - 36 -144
36(x-1)^2 - 9(y+2)^2 - 16(z-3)^2 = 144
divide by 144
(x-1)^2 /4 - (y+2)^2 /16 - (z+3)^2 /9 = 1
centre is (1,-2,-3)
So this is a hyperboloid of 2 sheets right?? Thanks so much!
To put the given quadratic surface in standard form and identify its center, we need to complete the square for each variable.
Let's start with the x variable:
36x^2 - 72x = 9(4x^2 - 8x)
To complete the square, we take half of the coefficient of x and square it: (-8/2)^2 = 16.
So, we can rewrite the x terms as: 9(4(x^2 - 2x + 4)) = 36(x - 1)^2 - 36.
Similarly, let's complete the square for y:
-9y^2 - 36y = -9(y^2 + 4y)
Taking half of the coefficient of y and squaring it: (4/2)^2 = 4.
Rewriting the y terms: -9(y^2 + 4y + 4) = -9(y + 2)^2 + 36.
Now, let's complete the square for z:
-16z^2 + 96z = -16(z^2 - 6z)
Taking half of the coefficient of z and squaring it: (6/2)^2 = 9.
Rewriting the z terms: -16(z^2 - 6z + 9) = -16(z - 3)^2 + 144.
Combining the completed square terms, we have:
36(x - 1)^2 - 9(y + 2)^2 - 16(z - 3)^2 - 36 + 36 + 144 - 288 = 0.
Simplifying further:
36(x - 1)^2 - 9(y + 2)^2 - 16(z - 3)^2 - 144 = 0.
Now that the quadratic surface is in standard form, we can identify its center. The center is given by the opposite sign of the square term coefficient. In this case, the center is located at (1, -2, 3) since the signs are opposite to the square terms (x - 1)^2, (y + 2)^2, and (z - 3)^2.