one line is tangent to y=1/x at pt. A and at the same time tangent to y=x^2 at pt. B . What is the distance AB

Answer 1

Okay, let a represent a point for x for y=1/x.
And c represent a point for x for y=x^2..
And the derivative of both equations should be equal, so -1/a^2=2c.
And the slope between the 2 points should also equal the deriatives. So, (c^2-(1/a))/(c-a) = 2c.
Or, 2c^2-2ac=c^2-(1/a).
And since we know -1/a^2=2c, or c=-1/2a^2 ,
then substitute in c in previous equation, so 2/4a^4--2a/2a^2=1/4a^4-(1/a).
And moving terms around, 1/4a^4=-2/a.
So, a=-8a^4, or 1/a^3=-8.
So, a=-1/2.
And c=-1/(-1/2)^2=-2.

So, you find the distance between (-1/2,-2) and (-2,4).
So, the distance is sqr((4--2)^2+(-2--1/2)^2) = sqr(36+(9/4)) = 6.184658 approximately
Thanks, for pointing out error!
I fixed itThanks again!

Answer 2

i'm assuming that there is a typo blunders and it could have examine y = 4x - x^2 we are able to locate the slope of the tangent at (a million,3) via differentiating y = ... : dy/dx = 4 - 2x [dy/dx = 4x^(a million-a million) - 2*x^(2-a million) = 4x^0 - 2x = 4 - 2x] on the factor (a million,3) , x = a million subsequently dy/dx = 4 - 2(a million) = 2, the *slope*, or gradient, of the tangent line. [in case you're able to desire to locate additionally the equation of the tangent line you substitute into this type of equation of a in the present day line y - y_0 = m (x - x_0) , it is plenty extra effective for fixing this type of circumstances. subsequently on the factor (a million,3), x = a million and y =3 you recognize that the slope is dy/dx = 2 = m So y - 3 = 2 (x - a million) , y = 2x -2 +3 y = 2x + a million , the equation of the tangent to the factor (a million,3) ]