Sketch the surface z 3y2 2x2 The traces in the vertical pl

Sketch the surface z = 3y^2 - 2x^2. The traces in the vertical planes x = k are the parabolas z = which open upward. The traces in y = k are the parabolas z = which open downward. The horizontal traces are = k, a family of hyperbolas. We draw the family of traces in Figure 2, and we show how the traces appear when placed in their correct planes in Figure 3. In Figure 1 we fit together the terms to form the surface z = 3y^2 - 2x^2, a hyperbolic paraboloid. Notice that the shape of the surface near the origin resembles that of a saddle. This surface will be investigated further in a later section when we discuss saddle points.

Solution

Given z = 3y² - 2x²

First, x = k

z = 3y² - 2k²

(z + 2k²)/3 = y²

Which is a parabola in z-y plane which opens upward.

second,. y = k

   z = 3k² - 2x²

^.   z = -2x² + 3k²

. (z - 3k²)/2 = -x²

which is a parabola in z-x plane which open downward.

Third, to become a hyperbolic equation, z has to be constant.

So z = k

so that 3y² - 2x² = k could be an equation of hyperbola.