Tangents to Curves: Intersection in First Quadrant

[mstudent123]

Homework Statement

Is there anything special about the tangents to the curves xy=1 and (x^2) - (y^2) =1 at their point of intersection in the first quadrant.

The Attempt at a Solution

I know what the derivatives of both functions are and what they look like when graphed. But, I'm not sure what to do with it beyond that. For the first function I have that dy/dx=-1/x^2 and for the second I have dy/dx = x/sqrt(-1+x^2). What is different about these functions and what happens in the first quadrant?

 

[Mark44]

Homework Statement

Is there anything special about the tangents to the curves xy=1 and (x^2) - (y^2) =1 at their point of intersection in the first quadrant.

The Attempt at a Solution

I know what the derivatives of both functions are and what they look like when graphed. But, I'm not sure what to do with it beyond that. For the first function I have that dy/dx=-1/x^2 and for the second I have dy/dx = x/sqrt(-1+x^2). What is different about these functions and what happens in the first quadrant?

Well, according to the problem statement, the functions intersect there. Seems to me that it would be useful to find the point of intersection.

 

[mstudent123]

I have been trying to do that. However, I have not been able to find the intersection of the tangents and there is not intersection of the original functions.

 

[jbunniii]

I have been trying to do that. However, I have not been able to find the intersection of the tangents and there is not intersection of the original functions.

There are in fact two points of intersection. Why do you think there are none?

You need to find (x,y) that satisfies both xy = 1 and x^2 - y^2 = 1. How do you solve two equations with two unknowns?

 

[HallsofIvy]

You don't need to find where they intersect.

With xy= 1, xy'+ y= 0 so y'= -y/x. With x^2- y^2= 1, 2x- 2yy'= 0 so y'= x/y. Without worrying about what the "points of intersection" are, what is the product of thos two derivatives for a given (x, y)?

 

[mstudent123]

Thank you!