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

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.

• Is the surface that you’re talking about the level set $xy-z=0$, then? – amd May 20 at 21:19
• This problem isn’t really very different from the one in your previous question. The way to solve each of them is similar. – amd May 20 at 21:34

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: $$\begin {cases}x-a=0\\ ay-z=0 \end {cases}\quad\text {and}\quad \begin {cases}y-b=0\\ bx-z=0 \end {cases}.$$
• @amd Of course one should see the equations in combination with the previous ones: $y=b$ and $x=a$, respectively. Should I underline the point in the answer? – user May 20 at 21:28