Tangent Points 5A7B4B

1. **State the problem:** We have a curve defined by $$y = 3x + \frac{12}{x}$$ passing through points A$(2, 12)$ and B$(6, 20)$. We want to find points C and D on the curve where the tangent is parallel to the line AB, and then find the equation of the perpendicular bisector of segment CD. 2. **Find the slope of line AB:** The slope $m_{AB}$ is given by $$m_{AB} = \frac{y_B - y_A}{x_B - x_A} = \frac{20 - 12}{6 - 2} = \frac{8}{4} = 2.$$ 3. **Find the derivative of the curve to get the slope of the tangent:** Given $$y = 3x + \frac{12}{x} = 3x + 12x^{-1},$$ we differentiate: $$\frac{dy}{dx} = 3 - 12x^{-2} = 3 - \frac{12}{x^2}.$$ 4. **Set the derivative equal to the slope of AB to find points C and D:** We want $$3 - \frac{12}{x^2} = 2,$$ which simplifies to $$3 - 2 = \frac{12}{x^2} \implies 1 = \frac{12}{x^2}.$$ 5. **Solve for $x^2$:** $$x^2 = 12,$$ so $$x = \pm \sqrt{12} = \pm 2\sqrt{3}.$$ 6. **Find the corresponding $y$ values for points C and D:** For $x = 2\sqrt{3}$, $$y = 3(2\sqrt{3}) + \frac{12}{2\sqrt{3}} = 6\sqrt{3} + \frac{12}{2\sqrt{3}} = 6\sqrt{3} + \frac{6}{\sqrt{3}}.$$ Simplify the second term: $$\frac{6}{\sqrt{3}} = 6 \times \frac{\sqrt{3}}{3} = 2\sqrt{3}.$$ So, $$y = 6\sqrt{3} + 2\sqrt{3} = 8\sqrt{3}.$$ For $x = -2\sqrt{3}$, $$y = 3(-2\sqrt{3}) + \frac{12}{-2\sqrt{3}} = -6\sqrt{3} - \frac{12}{2\sqrt{3}} = -6\sqrt{3} - 2\sqrt{3} = -8\sqrt{3}.$$ 7. **Coordinates of points C and D:** $$C = (2\sqrt{3}, 8\sqrt{3}), \quad D = (-2\sqrt{3}, -8\sqrt{3}).$$ 8. **Find the midpoint $M$ of segment CD:** $$M_x = \frac{2\sqrt{3} + (-2\sqrt{3})}{2} = 0,$$ $$M_y = \frac{8\sqrt{3} + (-8\sqrt{3})}{2} = 0,$$ so midpoint $M = (0,0).$ 9. **Find the slope of CD:** $$m_{CD} = \frac{-8\sqrt{3} - 8\sqrt{3}}{-2\sqrt{3} - 2\sqrt{3}} = \frac{-16\sqrt{3}}{-4\sqrt{3}} = 4.$$ 10. **Find the slope of the perpendicular bisector:** The perpendicular slope is the negative reciprocal: $$m_{\perp} = -\frac{1}{4}.$$ 11. **Write the equation of the perpendicular bisector passing through midpoint $M(0,0)$:** Using point-slope form: $$y - 0 = -\frac{1}{4}(x - 0) \implies y = -\frac{1}{4}x.$$ **Final answers:** - Points $C$ and $D$ are $$\boxed{(2\sqrt{3}, 8\sqrt{3}) \text{ and } (-2\sqrt{3}, -8\sqrt{3})}.$$ - Equation of the perpendicular bisector of $CD$ is $$\boxed{y = -\frac{1}{4}x}.$$