# What is the slope of the tangent line of (1-x)(4-y^2)-1/lny = C , where C is an arbitrary constant, at (1,2)?

Jan 9, 2017

It is not possible to specify a value for C. See graph, with the line y = 2. and the explanation.

#### Explanation:

C is seemingly single-valued, but really not so.

It is so, upon setting x = 1 and y = 2 in the equation.

Here, C is shown as -1 /ln 2.

With this C for the graph ,the line y = 2 meets it at point(s) given by

${\left(x - 1\right)}^{2} \left(0\right) = 0.$

The solution is x is arbitrary.

An attempt to find the equation to the virtual tangent at (1, 2) would

result in y = 2.

See the (y=2)-inclusive graph.

graph{(y-2+10^(-10)x)((x-1)(y^2-4)-1/ln y+1/ln 2)=0 [-10, 10, -5, 5]}

Jan 9, 2017

The slope is $0$.

#### Explanation:

Use implicit differentiation (or partial derivatives) to find

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{4 - {y}^{2}}{2 x y - 2 y + \frac{1}{y {\left(\ln y\right)}^{2}}}$.

At $\left(1 , 2\right)$, we get

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{4 - {\left(2\right)}^{2}}{2 \left(1\right) \left(2\right) - 2 \left(2\right) + \frac{1}{2 {\left(\ln 2\right)}^{2}}} = 0$

Given that $\left(1 , 2\right)$ is on the curve, we can see that $C = - \frac{1}{\ln} 2$

Here are some of the curves in this family.

red dots $\text{ }$ $c = 0$
red dash $\text{ }$ $c = - 0.9102$
blue dash $\text{ }$ $c = - 1.4$
black dash $\text{ }$ $c = - 1.43$
solid black $\text{ }$ $c = - 1.442695$ (that is: $c \approx - \frac{1}{\ln} 2$)

And here are the same near $\left(1 , 2\right)$

The point of at which the two branches meet appears to be $\left(0.74 , 2\right)$

And if we continue with values of $c$ less than $- \frac{1}{\ln} 2$, the branches separate again, but differently.

blue dash $\text{ }$ $c = - 1.4$
solid black $\text{ }$ $c = - 1.442695$ (that is: $c \approx - \frac{1}{\ln} 2$)
red dash $\text{ }$ $c = - 1.5$
green dash $\text{ }$ $c = - 2$