The slope of the first equation is -1, so look for the points on the second equation where the slope is also -1. Then adjust a so the first is tangential to the second.

In two of your videos you prove i^i = e^(-Pi/2) and the ith root of i, I^(1/i) = e^(Pi/2). When I multiply i^i by i^(1/i) I get i. But when I multiple the representations e^(-Pi/2) and e^(Pi/2) I get e. So, does i = e? Where have I gone wrong?

The slope of the first equation is -1, so look for the points on the second equation where the slope is also -1. Then adjust a so the first is tangential to the second.

The two answers I got: a=-2,2.

I think the right answer is 2. x+y=a, x*y=1 x+1/x=ax x^2-ax+1=0 x=a/2+-sqrt (a/2)^2-1

Discriminant must be >=0 (a/2)^2-1=>0 a=2

x*y=1 has a slope of -1 at both (-1,-1) and (1,1). The slope of x+y=a is ALWAYS -1.

In order for x+y=a to to go through (-1,-1), a must be -2.

In order for x+y=a to go through (1,1), a must be 2.

In both cases, x+y=a is a tangent line to x*y=1 at these respective points.

Dear Louis, your result is correct. I made a mistake. a must be +2 and -2. The discriminant must be 0 and therefor there is two solutions +2 and -2.