Parabola

The parabola $P$ passes through the points $K=[4; 5]$, $L=[2; 1]$ and $M=[-1; 0]$ and its symmetry axis is parallel to the $x$ axis. Determine the distance between the focus and the vertex of the parabola.

David solved this problem using the following steps:

(1) Initially, he wrote down the equation of the parabola in vertex form. The axis of parabola $P$ is parallel to the $x$-axis, so its vertex form is: $$P:(y - v_2)^2 = 2p(x - v_1),$$ where $[v_1; v_2]$ are the coordinates of the vertex and $|2p|$ is the distance of the focus $F$ from the directrix of the parabola $P$, i.e. $$|VF|=|p|$$

(2) Next, David used the information that the parabola passes through the points $K=[4; 5]$, $L=[2; 1]$ and $M=[-1; 0]$:

\begin{aligned} K \in P:\quad (5 - v_2)^2 &= 2p(4 - v_1)\cr L \in P:\quad (1 - v_2)^2 &= 2p(2 - v_1)\cr M \in P:\quad (0 - v_2)^2 &= 2p(-1 - v_1) \end{aligned}

(3) Solving the above system of three equations with three variables, David determined the value of $p$. At first, he expanded the parentheses in each of the equations: \begin{aligned} 25 - 10v_2 + v_2^2 &= 8p - 2pv_1\cr 1 - 2v_2 + v_2^2 &= 4p - 2pv_1\cr v_2^2 &= -2p - 2pv_1\cr\hline \end{aligned}

Then, he substituted the value of $v_2^2$ obtained from the third equation, into the first two equations: \begin{aligned} 25 - 10v_2 - 2p - 2pv_1 = 8p - 2pv_1\cr 1 - 2v_2 - 2p - 2pv_1 = 4p - 2pv_1\cr\hline \end{aligned}

Subsequently, he added the expression $2pv_1 + 2p$ to both sides of both equations:

\begin{aligned} 25 - 10v_2 &= 10p\cr 1 - 2v_2 &= 6p\cr\hline \end{aligned}

Then, he divided the first equation by $-5$:

\begin{aligned} -5 + 2v_2 &= -2p\cr 1 - 2v_2 &= 6p\cr\hline \end{aligned}

Finally, by summing the two equations, he calculated the value of $p$. $$-4 = 4p\ \Rightarrow \ p = -1$$

(4) David concluded that $|VF|=|p|=|-1|=1$.

Is David's solution correct? If not, determine where David made a mistake.