Robert's answer involves picking any $v$ and $w$ in ${\mathbf R}^n$ and getting an identity of polynomials of degree at most 2 in $t$ away from where $F(v+tw) = 0$, which excludes at most two values of $t$. He equates the coefficients in the identity to get the desired formula.  

The same technique works for nondegenerate quadratic forms over any field with at least five elements (since a quadratic polynomial function on a field is completely determined away from two points if at least three points remain). Of course we can ignore fields of size 2 and 4, since that would be in characteristic 2, but what happens in the case of quadratic forms over a field of size 3?  Admittedly the question was being asked over the real numbers for the purpose of a linear algebra class, so the whole point of checking finite fields isn't essential, but it's worth noting that the result is still true.  I'll indicate the steps, but leave out computational details. 

We have two nondegenerate quadratic forms $Q_1$ and $Q_2$ on a finite-dimensional vector space $V$ over a field $K$ of characteristic not 2, and we assume the isotropy groups of $Q_1$ and $Q_2$ are the same.  We want to show if $Q_1(v) \not= 0$ that $Q_2(v) \not= 0$ and $Q_2(w) = cQ_1(w)$ for all $w \in V$, where $c = Q_2(v)/Q_1(v)$.

a)  Let $v$ in $V$ be nonzero, $L \colon V \rightarrow K$ be linear, and $B \colon V \times V \rightarrow K$ be a nondegenerate bilinear form.  If $B(w,v)L(w) = 0$ for all $w \in V$, then $L(w) = 0$ for all $w \in V$. (Hint: There is a basis of $V$ in the complement of any hyperplane.)

b) Let $B_1$ and $B_2$ be the symmetric bilinear forms associated to $Q_1$ and $Q_2$. Suppose $v \in V$ satisfies $Q_1(v) \not= 0$.  Then by part a and a reflection as in Robert's answer, $Q_1(v)B_2(v,w) = Q_2(v)B_1(v,w)$ for all $w \in V$.

c) Assume from now on that $Q_1$ and $Q_2$ have the same isotropy group. Using part b, for all $v \in V$ we have $Q_1(v) \not= 0$ if and only if $Q_2(v) \not= 0$, and when $Q_1(v) \not= 0$ we have $\{w : B_1(v,w) = 0\} = \{w : B_2(v,w) = 0\}$.`

d) If $Q_1(v) \not= 0$ and $B_1(v,w) \not= 0$ then $Q_2(w) = (Q_2(v)/Q_1(v))Q_1(w)$. (To prove this, by part c we can assume $Q_1(w) \not= 0$.) 

e) If $Q_1(v) \not= 0$ and $B_1(v,w) = 0$ then $Q_2(w) = (Q_2(v)/Q_1(v))Q_1(w)$.  (By part c, $B_2(v,w) = 0$. We have $B_1(v,v+w) = B_1(v,v) = Q_1(v) \not= 0$, so we can apply part d with $v+w$ in place of $w$.)