Question

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)`?

Answer 1

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

`(x-1)^2(0)=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]}

Answer 2

The slope is `0`.

Explanation:

Use implicit differentiation (or partial derivatives) to find

`dy/dx = (4-y^2)/(2xy-2y+1/(y(lny)^2))`.

At `(1,2)`, we get

`dy/dx = (4-(2)^2)/(2(1)(2)-2(2)+1/(2(ln2)^2)) = 0`

Given that `(1,2)` is on the curve, we can see that `C = -1/ln2`

Here are some of the curves in this family.

red dots `" "` `c=0`
red dash `" "` `c=-0.9102`
blue dash `" "` `c=-1.4`
black dash `" "` `c=-1.43`
solid black `" "` `c=-1.442695` (that is: `c~~-1/ln2`)

And here are the same near `(1,2)`

The point of at which the two branches meet appears to be `(0.74,2)`

And if we continue with values of `c` less than `-1/ln2`, the branches separate again, but differently.

blue dash `" "` `c=-1.4`
solid black `" "` `c=-1.442695` (that is: `c~~-1/ln2`)
red dash `" "` `c=-1.5`
green dash `" "` `c=-2`