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

Question

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?

Answer 1

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 $y = \frac{a}{(x+2)^2}$

Answer 2

Hint:   the equation of the intersection $\,\dfrac{a}{(x+2)^2} = y_1 + \dfrac{y_2-y_1}{x_2-x_1}(x-x_1)\,$ must have a double root in $\,x\,$ for it to be a point of tangency.