There are two tangents from the point $P = (\frac{1261}{20},-1)$ to the ellipse with equation $\frac{x^2}{169}+\frac{y^2}{25} = 1.$
Determine the coordinates of the points, $E$ and $F$, where the tangents touch this ellipse. All coordinates should be correct to two decimal places.
So far I have found the slope to be $\frac{-25x}{169y}$. Is it right to say the equation is $y+1 = \frac{-25x}{169y}\left(x-\frac{1261}{20}\right)$ ? I attempted to sub this equation of a line into the ellipse equation to get the points, but i still have two variables? All help appreciated.
No, that's not correct. When you say you've found the slope to be $\frac{-25x}{169y}$, you need to be much more precise. What you know is that if a line is tangent to the ellipse at a point $(a, b)$ (which is on the ellipse, of course), then the slope of that line is $\frac{-25a}{169b}$. That means the equation of the line is $$ y - b = \frac{-25a}{169b}(x-a). $$
Notice that I did not use "$(x, y)$" as the point of tangency to the ellipse, because $(x, y)$ is used to denote a generic point of the tangent line, and only one of those points is on the ellipse.
So right there is where your approach went off the rails.
Let me suggest a different starting point:
Construct a line from $(1261/20, -1)$ to an arbitrary point $(a, b)$ on the ellipse. Determine the slope of the line, and compare this to the slope of the ellipse at $(a, b)$. For certain values of $(a, b)$, these two will be equal, and for those points, you've got a tangent line rather than a secant line.