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I' m asked to find the plane equation for (x,y,z)-> 0,0,0.

I' m given the following function: $$f(x,y)=e^{x^{2}}+\ln(\frac{1}{xy})$$ Also the task specifies that the origin plane is parallel with the tangent plane and passes through the following point (x,y,z)->(1,1,e) (tangent plane)

I've found the tangent plane equation (after doing the partial derivatives) and the equation is :$$x(2e-1)+y-z-(e+2)=0$$ (Implicit form).

How do I find the plane passing through (0,0,0), origin, given that I have now the tangent plane equation? Can I just copy the coefficients and write 0 at the end since they are parallel? Like this perhaps: $$x(2e-1)+y-z=0$$

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    $\begingroup$ Yes. A plane $ax+by+cz+d=0$ is always parallel to $ax+by+cz=0$. $\endgroup$ – Bobson Dugnutt Sep 1 '16 at 17:38
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Your approch is right, and but I got different numbers: the plane tangent at $P=(1,1,f(1,1))=(1,1,e)$ should be $$F_x(P)(x-1)+F_y(P)(y-1)+F_z(P)(z-e)=0\quad \Rightarrow\quad (2e-1)x-y-z-(e-2)=0$$ where $F(x,y,z)=z-e^{x^2}+\ln(xy)$ and $(F_x,F_y,F_z)=(-2xe^{x^2}+1/x,1/y,1)$.

Hence the parallel plane through $(0,0,0)$ is $(2e-1)x-y-z=0$.

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  • $\begingroup$ Hmm, it seems that I've misscalculated something. Tnx for the answer! $\endgroup$ – eugene_sunic Sep 1 '16 at 21:03
  • $\begingroup$ Could you please explain how you go z to be: 2e+1? I just get e, by putting x=1 and y=1 to the given function, simple as that... $\endgroup$ – eugene_sunic Sep 1 '16 at 21:32
  • $\begingroup$ @wesewx Your given point is $(1,1,2e+1)$. Check your question. $\endgroup$ – Robert Z Sep 2 '16 at 5:03
  • $\begingroup$ @omg, I'm sorry for that. $\endgroup$ – eugene_sunic Sep 2 '16 at 7:23
  • $\begingroup$ @wesewx So the given point is $(1,1,e)$ and not $(1,1,2e+1)$. Do you confirm? $\endgroup$ – Robert Z Sep 2 '16 at 7:44

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