Saturday, July 7, 2012

Lines on a Hyperbolic Paraboloid

A ruled surface has the property that through every point on the surface there is a straight line lying on the surface. Cylinders and cones are ruled surfaces. Another example is a hyperbolic paraboloid (a saddle surface). In fact, a hyperbolic paraboloid contains a pair of straight lines through any given point on it. Find equations of a pair of lines through the point $(1,2,3)$ on the hyperbolic paraboloid $z=y^2-x^2$. Prove that these are the only two such lines.

Solution: Remember that a line in space can be given in parametric form $x=x_0+\alpha t$, $y=y_0+\beta t$, $z=z_0+\gamma t$, where $x_0$, $y_0$, $z_0$, $\alpha$, $\beta$, $\gamma$ are constants, and $t \in \mathbf{R}$.
Let $x=1+t$, $y=2+t$. Then \[ z=(2+t)^2-(1+t)^2=3+2t. \] So we have a line $(x,y,z)=(1+t,2+t,3+2t)$. Similarly, let $x=1+t$, $y=2-t$. Then \[ z=(2-t)^2-(1+t)^2=3-6t. \] So we have a line $(x,y,z)=(1+t,2-t,3-6t)$.
We will prove that these are the only two lines on $z=y^2-x^2$ through $(1,2,3)$. Assume that $x=1+\alpha t$, $y=2+\beta t$. Then \[ z=(2+\beta t)^2-(1+ \alpha t)^2=4+4 \beta t + \beta^2 t^2 -1- 2 \alpha t - \alpha^2 t^2. \] In order for us to have a line, the coefficient of $t^2$ must be zero. Hence $\alpha^2=\beta^2$, and $\beta= \pm \alpha$. For definiteness, assume that $\alpha=1$. Then $\beta=\pm 1$, and we have the two cases above.

We could make the same kind of argument to show that given any point on the surface $z=y^2-x^2$, there are exactly two lines through the point and lying on the surface. But there is a method that renders the computations even easier. We make the change of variables \[ x=X-Y, \quad y=X+Y, \quad z=4Z. \] Then \[ 4Z=(X+Y)^2-(X-Y)^2=X^2+2XY+Y^2-X^2-Y^2+2XY, \] and we obtain the particularly simple equation \[ Z=XY. \] The critical thing is that the change of variables is linear; lines are mapped to lines. In our problem, the point was $(1,2,3)$. Under the change of variables, this becomes $(\frac{3}{2},\frac{1}{2},\frac{3}{4})$. The lines that pass through this point and lie on the saddle surface have either $X$ or $Y$ constant: \[ (X,Y,Z)=\left(\frac{3}{2}+s,\frac{1}{2},\frac{3}{4}+\frac{1}{2}s\right) \quad \mbox{and} \quad (X,Y,Z)=\left(\frac{3}{2},\frac{1}{2}+t,\frac{3}{4}+\frac{3}{2}t\right). \] In general, the two lines passing through the point $(X_0,Y_0,Z_0)$ and lying on the surface $Z=XY$ are \[ (X,Y,Z)=(X_0+s,Y_0,Z_0+sY_0) \quad \mbox{and} \quad (X,Y,Z)=(X_0,Y_0+t,Z_0+tX_0). \] Thus we have two families of lines that rule the saddle surface. The lines in each family do not intersect because if two such lines intersected, the intersection point would be on three lines on the saddle surface. Furthermore, every pair of lines from different families intersect. For two such lines have equations \[ (X,Y,Z)=(X_1+s,Y_1,Z_1+sY_1) \quad \mbox{and} \quad (X,Y,Z)=(X_2,Y_2+t,Z_2+tX_2), \] and it is easy to see that the lines intersect at the point $(X_2,Y_1,X_2Y_1)$, when $s=X_2-X_1$ and $t=Y_1-Y_2$.

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