The matrix
\[\begin{pmatrix} \frac{4}{29} & -\frac{10}{29} \\ -\frac{10}{29} & \frac{25}{29} \end{pmatrix}\]corresponds to projecting onto a certain vector $\begin{pmatrix} x \\ y \end{pmatrix}.$  Find $\frac{y}{x}.$
So if we let $x=\begin{pmatrix} 1 \\ 0 \end{pmatrix}$, then we have $Mx = \begin{pmatrix} \frac{4}{29} & -\frac{10}{29} \\ -\frac{10}{29} & \frac{25}{29} \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} \frac{4}{29} \\ -\frac{10}{29} \end{pmatrix}$. Right, so we see that $M\begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} \frac{4}{29} \\ -\frac{10}{29} \end{pmatrix}$. So $\alpha\begin{pmatrix} x \\ y \end{pmatrix}=\begin{pmatrix} \frac{4}{29} \\ -\frac{10}{29} \end{pmatrix}$. And we want to solve for $\frac{y}{x}$. We can see that $\alpha x=\frac{4}{29}$ and $\alpha y=-\frac{10}{29}$. So $\frac{y}{x}=-\frac{10}{4}=-\frac{5}{2}$.

