\(\displaystyle z \ = \ f(x,y) \ = \ \frac{1}{7}(4x^2-2y^2) \ at \ point \ (-2,-1,2).\)
\(\displaystyle Now, \ we \ want \ to \ find \ an \ equation \ to \ the \ tangent \ plane \ to \ f(x,y) \ at \ the \ point \ (-2,-1,2).\)
\(\displaystyle f_x(x,y) \ = \ \frac{8x}{7} \ and \ f_y(x,y) \ = \ \frac{-4y}{7}.\)
\(\displaystyle f_x(-2,-1) \ = \ \frac{-16}{7} \ and \ f_y(-2,-1) \ = \ \frac{4}{7}\)
\(\displaystyle Hence, \ \frac{-16}{7}(x+2)+\frac{4}{7}(y+1)-(z-2) \ = \ 0.\)
\(\displaystyle = \ \frac{-16x}{7}-\frac{32}{7}+\frac{4y}{7}+\frac{4}{7}-z+2 \ = \ 0\)
\(\displaystyle = \ -16x-32+4y+4-7z+14 \ = \ 0, \ = \ -16x+4y-7z-14 \ = \ 0\)
\(\displaystyle Ergo, \ final \ answer \ is \ 16x-4y+7z+14 \ = \ 0, \ QED.\)
\(\displaystyle Note: \ Since \ the \ point \ is \ tangent \ to \ both \ the \ plane \ and \ the \ f(x,y), \ then \ an \ easy \ check\)
\(\displaystyle is \ to \ make \ sure \ the \ point \ is \ in \ both \ equations.\)
