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:


However, the solution says the answer is:


What am I doing wrong here?

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.