equation for tangent plane

[JJ007]

Find an equation for the tangent plane to the surface $\displaystyle f(x,y)=\frac{1}{7}(4x^2-2y^2)$at point (-2,-1,2)

$\displaystyle f_{x}=\frac{-12}{7}?$
$\displaystyle f_{y}=\frac{4}{7}?$

The multiple choice answers are mostly 16x-4y.... =0 so there's something I'm doing wrong.

 

[BigGlenntheHeavy]

$\displaystyle z \ = \ f(x,y) \ = \ \frac{1}{7}(4x^2-2y^2) \ at \ point \ (-2,-1,2).$

$\displaystyle Now, \ we \ want \ to \ find \ an \ equation \ to \ the \ tangent \ plane \ to \ f(x,y) \ at \ the \ point \ (-2,-1,2).$

$\displaystyle f_x(x,y) \ = \ \frac{8x}{7} \ and \ f_y(x,y) \ = \ \frac{-4y}{7}.$

$\displaystyle f_x(-2,-1) \ = \ \frac{-16}{7} \ and \ f_y(-2,-1) \ = \ \frac{4}{7}$

$\displaystyle Hence, \ \frac{-16}{7}(x+2)+\frac{4}{7}(y+1)-(z-2) \ = \ 0.$

$\displaystyle = \ \frac{-16x}{7}-\frac{32}{7}+\frac{4y}{7}+\frac{4}{7}-z+2 \ = \ 0$

$\displaystyle = \ -16x-32+4y+4-7z+14 \ = \ 0, \ = \ -16x+4y-7z-14 \ = \ 0$

$\displaystyle Ergo, \ final \ answer \ is \ 16x-4y+7z+14 \ = \ 0, \ QED.$

$\displaystyle Note: \ Since \ the \ point \ is \ tangent \ to \ both \ the \ plane \ and \ the \ f(x,y), \ then \ an \ easy \ check$

$\displaystyle is \ to \ make \ sure \ the \ point \ is \ in \ both \ equations.$