Find a curve \(C\subset\mathbb R^2\) through \((3,2)\) with the following property. For \((x_0,y_0)\in C\), let \(L(x_0,y_0)\) be the segment of the tangent line to \(C\) at \((x_0,y_0)\) that lies in the first quadrant. Require that every \((x_0,y_0)\in C\) be the midpoint of \(L(x_0,y_0)\).
Solution
1. Define\[ C=\left\{(x,y)\in\RR^2:x>0,\ xy=6\right\}. \]Then \(C\) is a smooth curve in the first quadrant and contains \((3,2)\).
The equation \(xy=6\) with \(x>0\) is the graph \[ y=\frac{6}{x}, \] so it defines a smooth curve in the first quadrant. Since \[ 3\cdot2=6, \] the point \((3,2)\) lies on \(C\).
2. If \((x_0,y_0)\in C\), then the tangent line to \(C\) at \((x_0,y_0)\) meets the positive coordinate axes at\[ (2x_0,0) \qquad\text{and}\qquad (0,2y_0). \]
Since \(y=6/x\) on \(C\), \[ y'(x_0) = -\frac{6}{x_0^2} = -\frac{y_0}{x_0}, \] where the last equality uses \(x_0y_0=6\). Thus the tangent line has equation \[ Y-y_0 = -\frac{y_0}{x_0}(X-x_0), \] or equivalently \[ Y = 2y_0-\frac{y_0}{x_0}X. \] Setting \(Y=0\) gives \(X=2x_0\), while setting \(X=0\) gives \(Y=2y_0\).
3. Every point \((x_0,y_0)\in C\) is the midpoint of \(L(x_0,y_0)\).
Both \(x_0\) and \(y_0\) are positive. By step 2, the tangent line has negative slope and intersects the coordinate axes at \((2x_0,0)\) and \((0,2y_0)\). Hence its portion in the first quadrant is exactly the segment joining these two points. Its midpoint is \[ \left( \frac{2x_0+0}{2}, \frac{0+2y_0}{2} \right) = (x_0,y_0). \] Thus \((x_0,y_0)\) is the midpoint of \(L(x_0,y_0)\).
4. Q.E.D.
Steps 1 and 3 show that the curve \[ \boxed{C=\left\{(x,y)\in\RR^2:x>0,\ xy=6\right\}} \] passes through \((3,2)\) and has the required midpoint property.