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Find an equation for the plane tangent to $2x^2+3y^2−z^2=4$ at $(1,1,−1)$

So I started by taking the partial derivative for each term.

$\frac{\partial}{dx}=4x$ $\Rightarrow$ $f_x(1)=4$

$\frac{\partial}{dy}=6y$ $\Rightarrow$ $f_y(1)=6$

$\frac{\partial}{dz}=-2z$ $\Rightarrow$ $f_z(-1)=2$

So setting up the equation for the plane:

$4(x-1)+6(y-1)+2(z+1)=4$

However, the solution says the answer is:

$z=−2(x−1)−3(y−1)−1$

What am I doing wrong here?

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  • $\begingroup$ The end of your equation should be = $0$ instead of $4$ $\endgroup$ – Nicholas Nov 11 '15 at 3:18
  • $\begingroup$ @Nicholas, what happens with the $4$ in the original equation? $\endgroup$ – hax0r_n_code Nov 11 '15 at 3:19
  • $\begingroup$ Since it is tangent at $(1,1,-1)$ , you can substitute that point into your equation of the tangent to find that it should = $0$ $\endgroup$ – Nicholas Nov 11 '15 at 3:20
  • $\begingroup$ @Nicholas I don't see how?.. 2(1)+3(1)-(1) = 4... where is he wrong? $\endgroup$ – Prakhar Londhe Nov 11 '15 at 4:13
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    $\begingroup$ @Nicholas .. sorry I got what you meant... he meant at the end of plane equation... didn't you? $\endgroup$ – Prakhar Londhe Nov 11 '15 at 4:16
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It should be $0$ instead of $4$; to see a why notice:

Let $C := \{ (x,y,z) \in \Bbb{R}^{3} \mid 2x^{2}+3y^{2}-z^{2} = 4 \}$; let $f: (x,y,z) \mapsto 2x^{2}+3y^{2}-z^{2}$ on $\Bbb{R}^{3}$. Then $f^{(-1)}\{ 4 \} = C$; but $\nabla f(x,y,z) = (4x, 6y, -2z)$ for all $(x,y,z) \in \Bbb{R}^{3}$, which is a vector normal to $C$ at $(x,y,z)$ for all $(x,y,z) \in C$; hence $\nabla f(1,1,-1) = (4,6,2)$ is a vector normal to $C$ at $(1,1,-1)$, and then the plane tangent to $C$ at $(1,1,-1)$ is the set of all points $(x,y,z) \in \Bbb{R}^{3}$ such that $$ \nabla f(1,1,-1)\cdot (x-1,y-1,z+1) = (4,6,2)\cdot (x-1,y-1,z+1) = 0. $$

You may try to do something with the last equality to get what you want.

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