Suppose that we have a quadric $Q$ in $\mathbb P^3$ defined by the equation $x^2+y^2=z^2+w^2$ and consider the point $p=[1:1:1:1]$ on $Q$. Compute the equation of all the lines through $p$ contained in $Q$.
Now there is a clear way to do this by consider first a linear change of coordinates such that we make our quadric equation become $xw=yz$ and use the Segre embedding of $\mathbb P^1\times\mathbb P^1$ to get the desired answer. I was wondering if there was a more direct strategy to find an answer.
As Sasha says in a comment, the best way to get the two lines is to intersect the quadric with its tangent plane at $[1:1:1:1]$. In the open $w=1$, we get the tangent plane $x+y-z-1=0$. In Macaulay2: