differentiation

[jeca86]

A tangent line is drawn to the hyperbola xy=c at a point P.
(a) show that the midpoint of the line segment cut from this tangent line
by the coordinate axes is P.
(b) show that the triangle formed by the tangent line and the coordinate
axes always has the same area, no matter where P is located on the
hyperbola.

 

[stapel]

(I will assume, in what follows, that we are considering only the first quadrant.)

The line equation, in more friendly form, is "y = c/x". Then y' = -c/x².

Let P be the point (p, c/p). (That is, P is the point where x has the fixed value p.) Then the slope of the tangent to that point is m = y' = -c/p².

Use the point P and the slope m to find the equation of the tangent line. Then find the x- and y-intercepts. Then find the midpoint. When you find that the midpoint is the same as P, you will have completed part (a).

For part (b), draw the triangle and note that the height is the y-intercept and the base is the x-intercept. What then is the area? Does the point P, the value x = p, matter at all?

Hope that helps a bit.

Eliz.

 

[Unco]

(a) show that the midpoint of the line segment cut from this tangent line
by the coordinate axes is P.

Differentiate xy=c implicitly to get dy/dx = -y/x.

Let P = (p,q)

Plug into the straight-line equation
y - q = (-q/p) (x - p)

Y-intercepts: x=0; X-intercepts: y=0. And you are practically done.