# Finding tangent planes to $B$

Let $B = \{(x,y,z)| Ax^2 + By^2 + Cz^2 = D \}$ and a plane $ax + by + cz = k$. Find all the parallel planes to the given plane that are also tangent to $B$.

Clearly the normal vectors of those planes are proportional to $(a,b,c)$, thus $$ax + by + cz + d=0.$$ And the tangent plane of $B$ is of the form $\nabla F(x_0, y_0, z_0) \cdot(x-x_0, x-y_0, z- z_0)= a(x-x_0)+b(y -y_0) + c(z-z_0)=0$.

But I'm not sure what I'm missing in order to find how $d$ is related to $k$? Would appreciate any hint.

• What is $F$ in this case. May 9 '17 at 21:36
• $F = Ax^2 + By^2 + Cz^2 - D$ May 9 '17 at 21:37
• Hint: write the equation of the plane as $(a,b,c)\cdot(x,y,z)=k$. This says that the dot product of every point on the plane with the normal vector $(a,b,c)$ is constant. Does that give you any ideas for how to find the constant term?
– amd
May 9 '17 at 22:43
• Apparently I can find the tangent point by $(2Ax_0, 2By_0, 2Cz_0)=(a,b,c)$, and then by plugging it into $x_0a + by_0 +cz_0 +d=0$, express/find $d$. But I don't get the "right" answer. May 11 '17 at 22:12

The normal vector to the tangent plane at the point $M(x_0,y_0,z_0)$ is $$\nabla F(M)=(\frac {\partial F}{\partial x}(M),\frac {\partial F}{\partial y}(M),\frac {\partial F}{\partial z}(M))$$
$$=(2Ax_0,2By_0,2Cz_0)$$
• I can equate this to $(a,b,c)$ and then find $(x_0, y_0, z_0)$, but plugging it into $ax_0 + by_0 + cz_0+d=0$ doesn't help me to find the connection to $k$. May 9 '17 at 21:54