Quadric Surfaces — Question 4

Section 14-4 · Calculus pages 2707–2708

Show solutionHide solution

Question 4

Analyze the hyperbolic paraboloid z=x24−y29.z=\frac{x^2}{4}-\frac{y^2}{9}.

Tasks

  1. Derive the principal vertical traces.

  2. Classify horizontal traces for z=cz=c.

  3. Find every line through the origin contained in the surface.

  4. Justify the term “saddle.”

Question 4 – Solution

Strategy Fix one coordinate at a time, then factor the level-zero equation.

Diagram shown in the original SVG worksheet.

Step 1: Vertical traces If y=0y=0, then z=x2/4z=x^2/4, an upward parabola. If x=0x=0, then z=−y2/9z=-y^2/9, a downward parabola. Their opposite concavities create the saddle.

Step 2: Horizontal traces At z=cz=c, x24−y29=c.\frac{x^2}{4}-\frac{y^2}{9}=c. For c>0c>0 the hyperbola opens in the xx-direction; for c<0c<0 it opens in the yy-direction.

Step 3: Lines through the origin A contained line (x,y,z)=t(a,b,c)(x,y,z)=t(a,b,c) must satisfy tc=t2(a2/4−b2/9)tc=t^2(a^2/4-b^2/9) for all tt, so c=0c=0 and a2/4=b2/9a^2/4=b^2/9. Factoring gives (x2−y3)(x2+y3)=0.\left(\frac x2-\frac y3\right)\left(\frac x2+\frac y3\right)=0. Thus the two lines are z=0,y=±3x/2\boxed{z=0,\ y=\pm 3x/2}. Direct substitution confirms every point on each line lies on the surface.