Tangential planes of $f(x,y) := (y^2-x)(y^2-2x) $ in $(-1,1)$ and $(-1,-1)$ Let $f:\mathbb{R^2} \to \mathbb{R}$ with $f(x,y) := (y^2-x)(y^2-2x) $
How can I find the function rules $\tau_{(-1,1)}(x,y)$ and $\tau_{(-1,-1)}(x,y)$ of the tangential planes on the graph of the function $f$ in the points $(-1,1)$ and $(-1,-1)$?
I know that the general equation of a plane $E$ is given by
$$E: z = z_0 + f_x(x_0,y_0) (x-x_0) + f_y(x_0,y_0) (y-y_0)$$
So first I have to find out the partial derivatives:
$$f_x = 4x-3y^2$$
$$f_y = 4y^3-6xy$$
For the point $(-1,1)$ I get:
$$f_x = 4\cdot (-1) -3\cdot 1^2 = -7$$
$$f_y = 4\cdot1^3 - 6\cdot(-1)\cdot1 = 2$$
For the point $(-1,-1)$ I get:
$$f_x = 4\cdot (-1) - 3\cdot (-1)^2 = -1$$
$$f_y = 4\cdot (-1)^3 - 6\cdot (-1)\cdot (-1)= -10$$
To get $z_0$ we also evaluate $f$ on the point $(-1,1)$:
$$z_0 = f(-1,1) = (1^2 - (-1)) \cdot ((1^2) - 2\cdot (-1)) = 4$$ and  on the point $(-1,-1)$:
$$z_0 = f(-1,-1) = ((-1)^2 - (-1)) \cdot ((-1)^2 - 2\cdot (-1)) = 6$$
Now I can use the formula for the tangential plane for the point  $(-1,1)$:
$$E: z = z_0 + f_x(x_0,y_0) (x-x_0) + f_y(x_0,y_0) (y-y_0) \\
 = 4 + (-7)\cdot(x-(-1)) + 2 \cdot (y-1) \\
 = 4-7x + 7 + 2y - 2 \\
 = -7x + 2y + 9$$
So the first tangential plane is $E: z = -7x + 2y + 9$
I continue with the formula for the tangential plane for the point $(-1,-1)$:
$$E: z = z_0 + f_x(x_0,y_0) (x-x_0) + f_y(x_0,y_0) (y-y_0) \\
= 6 + (-1) \cdot (x- (-1)) + (-10) \cdot (y- (-1)) \\
= 6 -x - 1 - 10 y - 10 \\
= -x - 10y - 5 $$
So the second tangential plane is $E: z = -x - 10y - 5$
Two questions:
- Is this correct?
- How can I find the intersecting line of those two tangential planes (or how can I show that those areas are parallel/matching?)
 A: Given $f(x,y) := (y^2-x)(y^2-2x)$ find the equations of the tangent planes at $(-1,1)$ and $(-1.-1)$.
Note to OP: You made errors in computing some of the constants.
\begin{eqnarray}
f_x(x,y)&=&-3y^2+4x\\
f_x(-1,1)&=&-7\\
f_x(-1,-1)&=&-7\\
f_y(x,y)&=&2y(2y^2-3x)\\
f_y(-1,1)&=&10\\
f_y(-1,-1)&=&-10\\
f(-1,1)&=&6\\
f(-1,-1)&=&6
\end{eqnarray}
\begin{eqnarray}
\tau_{(-1,1)}(x,y)&=&f(-1,1)+f_x(-1,1)(x+1)+f_y(-1,1)(y-1)\\
&=&6-7(x+1)+10(y-1)\\
&=&-7x+10y-11
\end{eqnarray}
\begin{eqnarray}
\tau_{(-1,-1)}(x,y)&=&f(-1,-1)+f_x(-1,-1)(x+1)+f_y(-1,-1)(y+1)\\
&=&6-7(x+1)-10(y+1)\\
&=&-7x-10y-11
\end{eqnarray}
Find the equation of the line of intersection of the two tangent planes.
First, rewrite the equations of the tangent planes in the following form:
\begin{eqnarray}
7x-10y+z&=&-11\\
7x+10y+z&=&-11
\end{eqnarray}
In this form we see that the normal vectors for the two planes are
\begin{eqnarray}
\mathbf{N}_{(-1,1)}&=&(7,-10,1)\\
\mathbf{N}_{(-1,-1)}&=&(7,10,1)
\end{eqnarray}
The cross-product of these two vectors (or any non-zero multiple of the cross-product) will be a direction vector of the line of their intersection.
$$ (7,-10,1)\times(7,10,1)=(-20,0,140) $$
So let the normal vector of the line of intersection be
$$ \mathbf{N}=(1,0,-7) $$
But we also need a point common to both planes. But clearly, $(0,0,-11)$ lies on both planes. Thus the equation of the line of intersection is
\begin{eqnarray}
(x,y,z)=(0,0,-11)+(1,0,-7)t
\end{eqnarray}
In symmetric form this is
$$ x=\frac{z+11}{-7},\quad y=0 $$
