<s>[INST] How do I find the points of tangency given a 2 variable function and a normal vector? I'm given the two variable function $f(x,y) = 4x^2+7y^2+5xy+14$ and asked to find the (2) points on the surface where the vector $6\hat i + 69\hat j + 3\hat k$ is normal to the tangent plane.  So far I have written a generalized linearization using the points $f(a,b)$: $$f(a,b) = 4a^2+7b^2+5ab+14$$ $$\Rightarrow z_0 = 4a^2+7b^2+5ab+14$$ $$ $$ $$\frac{\partial f}{\partial x} = 8x+5y$$ $$\Rightarrow x_0 = 8a+5b$$ $$ $$ $$\frac{\partial f}{\partial y} = 14y+5x$$ $$\Rightarrow y_0 = 14b+5a$$ For the following linearization: $$ $$ $$L(x,y) = 4a^2+7b^2+5ab+14+(8a+5b)(x-a)+(14b+5a)(y-b)$$ $$ $$ Rearranging this to be in the standard form of a plane: $$ $$ $$x(8a+5b)+y(14b+5a)-z = 4a^2+7b^2+5ab-14$$ $$ $$ If the goal is to find the two points where the given vector is normal to this plane I am at a loss. Any help from this point would be greatly appreciated.  Edit: Fixed an error in the partial derivatives.
 [/INST] $z = f(x,y) = 4x^2 + 7y^2+5xy + 14$
The normal to the tangent plane is along the gradient of $f$
$\nabla f = \begin{bmatrix} 8 x + 5 y \\ 14 y + 5 x \\-1 \end{bmatrix} $
If $\nabla f$ is parallel to $ (6, 69, 3) $, then their cross product is zero.
$ \bigg(\nabla f\bigg) \times (6, 69, 3) = \bigg( (14 y + 5x)(3) + 69 , 6 (-1) - 3 (8x + 5y) , (8x + 5y) (69) - 6(14 y + 5x) \bigg) $
The right hand side of the above simplifies to
$ (42 y + 15 x + 69, - 24 x - 15 y - 6, 522 x + 261 y  \bigg) = 0$
Factoring $3$
$ (14 y + 5 x + 23, - 8 x - 5 y - 2, 174 x + 87 y) = 0 $
Solve the first two equations
$ 5 x + 14 y = -23 $
$ -8 x - 5 y = 2  $
You get
$ x = \dfrac{ (-23)(-5) - (2)(14) }{-25 + 112} = \dfrac{ 87}{87} = 1 $
$ y = \dfrac{-8 - 2}{5} = -2$
Plug that into the third equation
$ (1)(174) + (87)(-2) = 0 $
So the solution of the first two equation satisfies the third equation.
The solution point is $ x = 1, y = -2, z = 4(1)^2 + 7(-2)^2 +5(1)(-2) + 14  = 36 $
And therefore, the point of tangency is $(1, -2, 36) $
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