# 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)#?

##### 2 Answers

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

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]}

The slope is

#### Explanation:

Use implicit differentiation (or partial derivatives) to find

At

Given that

Here are some of the curves in this family.

red dots

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And here are the same near

The point of at which the two branches meet appears to be

And if we continue with values of

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solid black

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