Equation of a tangent line to a curve Equation of the curve is $y=(x+9)/(x+5)$, we are looking for the tangent line to that curve that also goes through $O(0,0)$. Answer given is $x+25y=0$ which I found to be true for $A(-15, 3/5)$ being part of the line and the curve. Question is, how we got to that answer. After differentiating the curve equation, we get $y'=-4/(x+5)^2$ which afterwards gives the equation $4x+25y=0$. There is a $4$ that shouldn't be there, where is my mistake? Thanks in advance. 

This doesn't always work, but when it does, it means that you can find the equation of the tangent line without using calculus.
A line through the origin must have the form $y_{_L} = mx$. When does such a line intersect the graph of the equation $y_{_H}=\dfrac{x+9}{x+5}$?
\begin{align}
   y_{_L} &= y_{_H} \\
   mx &= \dfrac{x+9}{x+5} \\
   mx^2 + 5mx &= x+9 \\
   mx^2 +(5m-1)x - 9 &= 0 \\
   x &= \dfrac{1-5m \pm \sqrt{(5m-1)^2+36m}}{2m} \\
\end{align}
If the line $y_{_L}=mx$ were a tangent line, there would only be, locally, one solution for $x$. This will happen when the descriminant is equal to $0$.
\begin{align}
   (5m-1)^2+36m &= 0 \\
   25m^2 +26m + 1 &= 0 \\
   (25m+1)(m+1) &= 0 \\
   m &\in \left\{-\dfrac{1}{25}, -1 \right\}
\end{align}
So we find two lines: $y_{_\ell} = -\dfrac{1}{25}x$ and $y_{_u} = -x$.
We still need to justify that these are, indeed, tangent lines. The equation 
$y_{_H}=\dfrac{x+9}{x+5}$ describes a "tilted" hyperbola and it appears that the line 
$y_{_\ell}=-x$ is tangent to the upper branch while the line $y_{_u} = -\dfrac{1}{25}x$ is tangent to the lower branch of the hyperbola.
In the first case, we find 
$y_{_H} - y_{_\ell} = 
   \dfrac{x+9}{x+5} + x =
   \dfrac{(x+3)^2}{x+5}$
We see that, near $x=-3$, this is approximately the parabola $y =\dfrac 12(x+3)^2$, which is tangent to the $x-$axis, from above, at $x=-3$. In other words, the two curves are tangent to each other at $x=-3$
In the other case, we find 
$y_{_u} - y_{_H} = 
   -\dfrac{x}{25} - \dfrac{x+9}{x+5} =
   -\dfrac{(x+15)^2}{25(x+5)}$
We see that, near $x=-15$, this is approximately the parabola 
$y =-\dfrac 52(x+15)^2$, which is tangent to the $x-$axis, from below, at $x=-15$. In other words, the two curves are tangent to each other at $x=-15$
