About a tangent line on the curve

Question

The curve is given by xy^3+2y-x^2+3x=-2. If y is differentiable with respect to x then the slope of the tangent line at the point (x_1,y_1) is given by m=y'(x_1) so that the equation of this tangent line is y-y_1=m(x-x_1).

For instance, if (x_1,y_1)=(2,-1) then I was able to compute manually that the tangent line is $-x+4y=-6$. When I tried to solve the problem using GeoGebra, there were two answers provided by GeoGebra. Of course, happy to see that one of the two answers is -x+4y=-6. Now, what made me distracted is the other answer given by 1.58x-1.99x=5.14. I really do not know the idea behind this answer.

Well, I tried to think that x is a differentiable with respect to y and started solving for x'(y) implicitly but unfortunately not able to get the answer 1.58x -1.99y=5.14.

Answer 1

Your curve is not a function, but an implicit curve. This means you are using this overload of the Tangent command:

Tangent( <Point>, <Implicit Curve> )

Creates the tangent to the implicit curve in the given point.

Example: Tangent((1,1), x^2+y^2=1)) yields lines x=1 and y=1.

Although it is not super clear from that description, GeoGebra generates all tangents passing through the given point, not just the tangent at that point (the example provided does support that).

This behavior is more appearent when the point is not on the curve:

So, the other solution you see is actually the tangent at around x=3.98.

These tangents can be given by the equation of a line , but with m=y'(x), instead of m=y'(x_0).

Differentiating yields

.

Then solving the system of equations

...does indeed result in x=2 and x=3.98651