Question

Reduce the equation to one of the standard forms in order to classify the surface.

z²+y²-4y-x²+3=0

A. Hyperboloid of one sheet with center (0, 2, 0)
B. Ellipsoid with center (0, 2, 0)
C. Cone with vertex (0, 2, 0)
D. Elliptic paraboloid with vertex (2, 0, 0)
E. Hyperboloid of two sheets with center (2, 0, 0)

Answer 1

The given equation can be transformed into the standard form of a hyperboloid of one sheet with center (0, 2, 0), identifying option A as the correct classification of the surface.

Step-by-step explanation:

The equation provided is z²+y²-4y-x²+3=0, which we need to transform into one of the standard forms to classify the surface. First, let's rearrange the terms related to each variable and complete the square for the y-terms:

z² + (y² - 4y + 4) - x² - 1 = 0

Add 1 to both sides to balance the equation:

z² + (y - 2)² - x² = 1

In this form, we can see the coefficients of y and z are positive and identical, while the coefficient of x is negative. This is indicative of a hyperboloid of one sheet, with its center at the point where y and z are zero (since we did not shift these variables) and where x terms complete the square inverse, which is still at zero. Hence, the center is (0, 2, 0).

Therefore, the correct answer is A. Hyperboloid of one sheet with center (0, 2, 0).