Which are the common points of a cone and the tangent hyperplane to this cone? Let $A$ be an invertible, symmetric, $n\times n$ real matrix,   $u\in \mathbb R^n$, $u\neq 0$, be such that $u^TAu=0$. 
Let us consider the following system of equations
$
x^TAx=0, 
u^TAx=0.
$
Geometrically , its solution are the common points of the cone $x^tAx=0$ and the tangent hyperplane to this cone at the point $u$ (from this cone).
Clearly, the points of the form $x=tu$, $t\in \mathbb R$ satisfy this system.
Are there another solutions?  
 A: In general, YES, there are others. As an example, take
$$A= diag(1, -1,\; 1,-1),\; u=(1,1,0,0)^T,\; x= (0,0,1,1)^T.$$ It is easy to check that the equations are satisfied and that $x$ and $u$ are not parallel. Here is how I found this counterexample:
Take any $x \in \Bbb R^n$. Let $$\{\lambda_1,\ldots,\lambda_n\}=\sigma(A),$$ that is, the set of eigenvalues of $A,$ and let 
$$\{v_1,\ldots,v_n\}$$ be a ortonormal basis of eigenvectors, that is, 
$$\|v_i\|=1,\; v_i^Tv_j=0 \textrm{ if } i\neq j,\; Av_i=\lambda_i v_i.$$ This vectors are well defined since $A$ is symetric and hence diagonalizable. Since the $v_i$'s are a basis of $\Bbb R^n,$ there exists scalars $\alpha_i,\beta_i$ for $i=1,\ldots,n$ such that:
$$u= \sum_{i=1}^n \alpha_i v_i,\; x=\sum_{i=1}^n \beta_i v_i.$$ The conditions $$u^TAu= x^TAx= u^TAx=0$$ are now equivalents to
$$0=\sum_{i=1}^n \alpha_i^2 \lambda_i=\sum_{i=1}^n \beta_i^2 \lambda_i=\sum_{i=1}^n \alpha_i\beta_i \lambda_i.$$ Given the values of $\alpha_i,$ we see that $x$ satisfies the equations if and only if the $\beta_i$'s satisfies the above system, in which we have 2 equations and $n$ variables. This made me realize that there are indeed another vectors $x$ that do not lie on $span\{u\}.$
Hope this helps. 
