I've come across a problem that I cannot finalize.
Problem: In which points the tangent plane of the function
$$ f(x,y) = 5 + \frac{x^2}{16} + \frac{y^2}{9} $$
is parallel to the plane given by the equation:
$$ z = x + 2y \ ? $$
I can compute the partial differentiations:
$$ \frac{\partial z}{\partial x}' = \frac{x}{8}, $$
$$ \frac{\partial z}{\partial y}' = \frac{2y}{9}, $$
so that the tangent plane has equation $$ z = \frac{\partial z}{\partial x}(x, x_0)(x-x_0) + \frac{\partial z}{\partial y}(y, y_0)(y-y_0) + z_0. $$
That gives us the tangent plane:
$$ z = \frac{-9 x_0^2 – 16 y_0^2 + 720}{144} + \frac{x x_0}{8} + \frac{2 y y_0}{9} $$
I cannot see how can I find the point where the tangent plane is parallel to $z$?
Thank you for any clue.
Best Answer
After all the hints and a couple of hours, I was finally able to figure it out.
With the tangent plane equation, we just need to look for the "linear combination" of the plane, meaning:
$$ z = \frac{\partial z}{\partial x}(x,x_0)(x-x_0) + \frac{\partial z}{\partial y}(y,y_0)(y-y_0) + z_0 $$
needs to match the coeficient of the normal vector given by the paralel plane $$ z = x + 2y$$ which has the normal vector $(1,2,-1)^T$.
That gives us the following equations: $$ \frac{x_0}{8} = 1$$ $$ \frac{2y_0}{9} = 2$$
which give us solutions: $x_0 = 8, y_0 = 9$. After the substitution we get the following formula:
$$ x + 2y - 18 - z + z_0 = 0 $$
Value of $z_0$ can be obtained from the assignment formula, ergo:
$$ z_0 = 5 + \frac{x_0^2}{16} + \frac{y_0^2}{9}, $$
$$ z_0 = 5 + \frac{64}{16} + \frac{81}{9}, $$
$$ z_0 = 5 + 4 + 9, $$
$$ z_0 = 13. $$
The desired tangent plane has the following formula: $$ z = x + 2y - 8. $$
Thanks for all your comments.