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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.

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  • $\begingroup$ Is the surface that you’re talking about the level set $xy-z=0$, then? $\endgroup$ – amd May 20 at 21:19
  • $\begingroup$ This problem isn’t really very different from the one in your previous question. The way to solve each of them is similar. $\endgroup$ – amd May 20 at 21:34
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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: $$ \begin {cases}x-a=0\\ ay-z=0 \end {cases}\quad\text {and}\quad \begin {cases}y-b=0\\ bx-z=0 \end {cases}. $$

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  • $\begingroup$ Those last two equations are those of planes, not lines. $\endgroup$ – amd May 20 at 21:21
  • $\begingroup$ @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? $\endgroup$ – user May 20 at 21:28
  • $\begingroup$ Absolutely, since that’s not at all what you wrote: “... the intersection lines are ...” followed by equations of two planes. $\endgroup$ – amd May 20 at 21:31
  • $\begingroup$ @amd Please check my edit. $\endgroup$ – user May 20 at 21:36
  • $\begingroup$ Looks good to me. $\endgroup$ – amd May 20 at 21:37

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