Quadric Surfaces — Question 4

Question 4 – Solution

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

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

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

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