Show that the tangent plane of the saddle surface $z=xy$ at any point intersects the surface in a pair of lines.

Question

My attempt: Let $f(x,y,z)=xy-z$ , (a,b,c)$\in$the saddle surface, and calculate the total derivative $Df(a,b,c)=(b,a,-1)$ Then the tangent plane is $$g(x,y,z)=f(a,b,c)+Df(a,b,c)(x-a,y-b,c-z)=bx+ay-z+ab$$
Set $g(x,y,z)=f(x,y,z)$ to get the intersection got $bx+ay-xy+ab=0$.
I know that the equation can be written as $$bx+ay-xy+ab=bx-ay-z-c=0$$ But I have no idea to get a pair of lines which intersects the saddle surface.

Answer 1

First of all you made an error in the calculation: $$\begin{align} g(x,y,z)=f(a,b,c)+Df(a,b,c)(x-a,y-b,\color{red}{z-c})&=bx+ay-z\color{red}{+c-2ab}\\ &=bx+ay-z\color{red}{-ab} \end{align} $$ where we used $ab-c=0$.

Now the intersection of the surfaces can be found from the equation: $$ xy-z=bx+ay-z-ab\implies (x-a)(y-b)=0, $$ which solutions are $x=a$ and $y=b$.

Substituting the values into equation of any of two surfaces one obtains that the intersection lines are: $$ \begin {cases}x-a=0\\ ay-z=0 \end {cases}\quad\text {and}\quad \begin {cases}y-b=0\\ bx-z=0 \end {cases}. $$