Posted by Jack on October 22, 2002 at 08:51:21:
In Reply to: can anyone give me a hand posted by skyparker on October 21, 2002 at 18:56:38:
: Here's the question:
: The tangent to the graph of 5xy=k at x=a is parallel to the secant through the points where x=b and x=c. Show aČ = bc.
: Rewrite y=k/5x
: The slope of the tangent line is y'=-k/(5x^2)
: At x=a, y'(x=a)=-k/(5a^2)
: The slope of the secant line through x=b and x=c is [y(c)-y(b)]/[c-b] = [k/(5c)-k/(5b)]/[c-b]
: = -k/(5cb)
: Since the tangent line and secant are parallel their slopes are equal. So
: -k/(5cb) = -k/(5a^2) or a^2=cb