A curve in the first quadrant has this property…

Step 1. The section formula. $P = \dfrac{2A + B}{3}$, so $x = \dfrac{2}{3}\left(x - \dfrac{y}{y'}\right)$ and $y = \dfrac{1}{3}(y - xy')$.

Step 2. Simplify. The first gives $3x = 2x - \dfrac{2y}{y'}$, so $x = -\dfrac{2y}{y'}$ and $y' = -\dfrac{2y}{x}$. The second gives $2y = -xy'$, which is the same equation.

Step 3. Solve. $\dfrac{dy}{y} = -\dfrac{2\,dx}{x}$ gives $\ln y = -2\ln x + \text{constant}$, so $x^2y = c$.

Step 4. The constant. $2^2 \times 3 = 12$, so $c = 12$.

Step 5. The height and the slope. At $x = 4$: $y = \dfrac{12}{16} = 0.75$. At $(2, 3)$: $y' = -\dfrac{2 \times 3}{2} = -3$.