"Find all plane curves in the first quadrant such that for every point
$P$ on the curve, $P$ bisects the part of the tangent line at $P$ that lies
in the first quadrant."
I found that the tangent line at $P$ must intersect the positive $x$– and $y$-axes. Also, the slope of the tangent at such a point is necessarily negative. Let the slope at this $P(x_0,y_0)$ is $m$ so that the tangent line is given by $$y-y_0=m(x-x_0).$$
I know how to find the $x$– and $y$-intercepts of this line but I can't manage to get anything after finding them, I appreciate any help, and thanks in advance.
Best Answer
Let $P$ be $(x,y)$, The eq. of atangent at P is $$Y-y=m (X-x), m=\frac{dy}{dx}$$ The co-ordinates of the points of intersction at x and y axis are $A(x-y/m,0), B(0,y-mx)$, the mid point of $AB$ is $P$ so we have $$x=\frac{x-y/m}{2} \implies \frac{dy}{dx}=-\frac{y}{x} \implies \int \frac{dy}{y}=- \int \frac{dx}{x} \implies \ln y=-\ln x+\ln C \implies xy=C$$ All such curves rectangular hyperbolas: $xy=C$