Whether or not a plane is a tangent plane to an ellipse or not? And if so, what is the point of intersection Say we have an ellipse 

Transpose( p-c )A(p-c) = 1

and a plane

x = a

where Transpose (p-c) implies the transpose of the array p-c and a is a const
A is an nxn matrix where the ellipse has n dimensions
Is there a way to check if the ellipse meets the plane at a single point? and if yes, what is that single point
P.S. If anyone can solve it for a sphere (3 dimensions) I will be equally content and can extend it to solve my problem
Edit:
A is the positive definite matrix. c is its center. p is any coordinate (variable in n dimensions) and x = a is an equation of the plane, whose forst coordinate is the constant, a.
Thanks
 A: $\newcommand{\Basis}{\mathbf{e}}\newcommand{\Reals}{\mathbf{R}}$Let $A$ be a positive-definite $n \times n$ real matrix, $c = (c_{1}, \dots, c_{n})$ a point of $\Reals^{n}$, $p = (p_{1}, \dots, p_{n})$ Cartesian coordinates, and $Q:\Reals^{n} \to \Reals$ the (positive-definite) quadratic form
$$
Q(p) = (p - c)^{t}A(p - c).
$$
Theorem: The ellipsoid $\{Q(p) = 1\}$ is tangent to the hyperplane $\{p_{1} = a\}$ if and only if the upper-left entry of $A^{-1}$ is $(a - c_{1})^{2}$, and in this event the point of tangency is
$$
p = c + \frac{1}{a - c_{1}} A^{-1} \Basis_{1}.
$$
Proof: The gradient of $Q$ is the vector field
$$
\nabla Q(p) = 2A(p - c).
$$
Particularly, the gradient is parallel to the first coordinate direction if and only if there is a real number $\lambda$ such that
$$
2A(p - c) = \lambda\Basis_{1}.
$$
The ellipsoid $\{Q(p) = 1\}$ is tangent to the hyperplane $\{p_{1} = a\}$ if and only if there is a $p$ satisfying the system:
\begin{align*}
&(p - c)^{t}A(p - c) = 1 &&\text{$p$ lies on the ellipsoid,}
\tag{1} \\
&2A(p - c) = \lambda\Basis_{1} && \text{ellipsoid tangent to $\{p_{1} = \text{const}\}$ at $p$,}
\tag{2} \\
&(p - c)^{t}\Basis_{1} = a - c_{1} && \text{first coordinate of $p$ is $a$.}
\tag{3}
\end{align*}
(Such a $p$ is necessarily unique since the ellipsoid is strictly convex.)
Substituting (2) into (1) gives
$$
\tfrac{1}{2} \lambda (p - c)^{t} \Basis_{1} = 1.
$$
Comparing with (3),
$$
\tfrac{1}{2} \lambda (a - c_{1}) = 1.
\tag{4}
$$
Solving (2) for $p - c$ and using (4) to eliminate $\lambda$,
$$
p - c
  = \tfrac{1}{2} \lambda A^{-1} \Basis_{1}
  = \frac{1}{a - c_{1}} A^{-1} \Basis_{1}.
\tag{5}
$$
Transposing and multiplying by $\Basis_{1}$ (i.e., dotting both sides with $\Basis_{1}$), using (3), and rearranging gives, finally,
$$
(a - c_{1})^{2} = \Basis_{1}^{t} A^{-1} \Basis_{1} = (A^{-1})_{11};
$$
that is, the upper-left entry of $A^{-1}$ is $(a - c_{1})^{2}$. The point of tangency is read off (5).

This reduces correctly if $A = \frac{1}{r^{2}}I$ is the scalar matrix whose "unit ellipsoid" is a sphere of radius $r$: The hyperplane $\{p_{1} = a\}$ is tangent to the sphere of radius $r$ centered at $c$ if and only if $c_{1} + r = a$, i.e., $r^{2} = (a - c_{1})^{2}$, and this is the upper-left entry of $A^{-1}$.
