calculus problem, can anyone answer this?

Question

find the curve that passes through the point (3,2) and has the property that if the tangent line is drawn at any point P on the curve, then the part of the tangent that lies in the first quadrant is bisected at P.

Answer 1

Consider the tangent line at a point P = (x, y). It has form y = mx + b (where m = y'(x)), and contains points (0, b) and (-b/m, 0). For the question to make sense, b and -b/m must both be positive. Then these points will be the endpoints of the line segment that is the part of the tangent line in the first quadrant. So the bisection point is (-b/2m, b/2) and this must be P = (x, y). So we have, for any x and y on the curve, -b/2m = x and b/2 = y (where b, m and y are all functions of x). Hence we get y'(x) = m = -y/x.

This is a separable ODE: we can rewrite it as
y' / y = -1 / x
=> (dy/dx) / y = -1/x
and hence
int (dy / y) = int (- dx / x)
i.e. ln |y| = - ln |x| + c
As the first quadrant is specified, we get y = exp(- ln x + c) = k / x where k > 0.

In this case the curve contains (3, 2) so k = 6 and the equation is

y = 6/x, x > 0.

Answer 2

By inspection I see that the curve that fits this description would have a slope of -y/x at any point on the curve.

The curve y = 1/x has the right slope but doesn't pass thru the point (3,2). It passes thru (3,1/3). The y value needs to be six times greater. So try

y = 6/x

Checking the slope:

dy/dx = -6/x^2 = -6/[x(6/y)] = -y/x as required.

y = 6/x is confirmed.

Answer 3

how can the tangent drawn at point P be bisected at point P.. there must be a problem in your question... check that...