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. 
 A: Let the $x$-coordinate of the point of tangency be $a$. By your calculation of the derivative, the tangent line when $x=a$ has slope $-\dfrac{4}{(a+5)^2}$.
But the line passes through the origin and $\left(a,\frac{a+9}{a+5}\right)$. It follows that
$$\frac{-4}{(a+5)^2}=\frac{\frac{a+9}{a+5}}{a}.$$
Manipulation gives $-4a=(a+9)(a+5)$, which has the solutions $a=-3$ and $a=-15$.
Note that at the point where $a=15$, your expression for the slope gives $-\frac{4}{(-10)^2}$, that is, $-\frac{1}{25}$. So the equation $x+25y=0$ is correct.
A: 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 $x=-15$
