Given two points that are joined by a line that is a tangent to a curve, find the missing constant in the equation for the curve How do I solve a question like this?
I have been given two points that make up a line tangent to $y=\frac{a}{(x+2)^2}$. I need to determine $a$.
First, I calculated the equation of the tangent line with $\frac{y2-y1}{x2-x1}$, and also found $\frac{dy}{dx}$. I had planned to set the equal derivative to the slope of the tangent and plug in an $x$ value from one of the points, however they both got a different result.
Is there a step I am missing?

The tangent point and slope there are the same for both equations.
Therefore $mx + c = \frac{a}{(x+2)^2}$ and $m = -\frac{2a}{(x+2)^3}$ for a
certain $x$ value.
To find the x value, set $(mx + c)(x+2)^2 = a$
And $m(x+2)^3 = -2a$
So $-\frac{m}{2}(x+2)^3 = (mx + c)(x+2)^2$
$-\frac{m}{2}(x+2) = mx + c$
$-\frac{mx}{2} - m = mx + c$
$\frac{3mx}{2} = -m - c$
$3mx = -2m - 2c$
$x = -\frac{2m+2c}{3m}$
Once you have $x$, find $y$ from $y = mx + c$ and then find $a$ from $y = \frac{a}{(x+2)^2}$
