1. **State the problem:** We have a curve $C$ defined by the equation $$x = \frac{2}{3} \sin(3y + \frac{\pi}{4})$$ with domain $$\frac{\pi}{12} < y < \frac{3\pi}{4}$$. We need to: (a) Find the exact $y$-coordinates of points $A$ and $B$ where the curve intersects the $y$-axis (i.e., where $x=0$). (b) Show that $$\left(\frac{dy}{dx}\right)^2 = \frac{1}{p - qx^2}$$ for some integers $p$ and $q$. (c) Using the result from (b) and the sketch, find the exact $x$-coordinate of point $D$, the intersection of the normal to $C$ at $A$ and the tangent to $C$ at $B$. --- 2. **Part (a): Find $y$ at points $A$ and $B$ where $x=0$** Since $A$ and $B$ lie on the $y$-axis, $x=0$. Set $$0 = \frac{2}{3} \sin(3y + \frac{\pi}{4}) \implies \sin(3y + \frac{\pi}{4}) = 0.$$ The general solution for $\sin \theta = 0$ is $$\theta = n\pi, \quad n \in \mathbb{Z}.$$ So, $$3y + \frac{\pi}{4} = n\pi \implies 3y = n\pi - \frac{\pi}{4} \implies y = \frac{n\pi}{3} - \frac{\pi}{12}.$$ We must find $n$ such that $$\frac{\pi}{12} < y < \frac{3\pi}{4}.$$ Try $n=1$: $$y = \frac{\pi}{3} - \frac{\pi}{12} = \frac{4\pi}{12} - \frac{\pi}{12} = \frac{3\pi}{12} = \frac{\pi}{4}.$$ Try $n=2$: $$y = \frac{2\pi}{3} - \frac{\pi}{12} = \frac{8\pi}{12} - \frac{\pi}{12} = \frac{7\pi}{12}.$$ Both $\frac{\pi}{4}$ and $\frac{7\pi}{12}$ lie in the domain. Therefore, - Point $A$: $$y = \frac{\pi}{4}$$ - Point $B$: $$y = \frac{7\pi}{12}$$ --- 3. **Part (b): Show that $$\left(\frac{dy}{dx}\right)^2 = \frac{1}{p - qx^2}$$** Given: $$x = \frac{2}{3} \sin(3y + \frac{\pi}{4}).$$ Differentiate both sides with respect to $y$: $$\frac{dx}{dy} = \frac{2}{3} \cdot \cos(3y + \frac{\pi}{4}) \cdot 3 = 2 \cos(3y + \frac{\pi}{4}).$$ Then, $$\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} = \frac{1}{2 \cos(3y + \frac{\pi}{4})}.$$ Square both sides: $$\left(\frac{dy}{dx}\right)^2 = \frac{1}{4 \cos^2(3y + \frac{\pi}{4})}.$$ Use the Pythagorean identity: $$\sin^2 \theta + \cos^2 \theta = 1 \implies \cos^2 \theta = 1 - \sin^2 \theta.$$ Recall from the original equation: $$x = \frac{2}{3} \sin(3y + \frac{\pi}{4}) \implies \sin(3y + \frac{\pi}{4}) = \frac{3x}{2}.$$ Therefore, $$\cos^2(3y + \frac{\pi}{4}) = 1 - \left(\frac{3x}{2}\right)^2 = 1 - \frac{9x^2}{4} = \frac{4 - 9x^2}{4}.$$ Substitute back: $$\left(\frac{dy}{dx}\right)^2 = \frac{1}{4 \cdot \frac{4 - 9x^2}{4}} = \frac{1}{4 - 9x^2}.$$ Hence, $$\left(\frac{dy}{dx}\right)^2 = \frac{1}{4 - 9x^2}$$ with integers $p=4$ and $q=9$. --- 4. **Part (c): Find the $x$-coordinate of point $D$** - Point $A$ has coordinates $(0, \frac{\pi}{4})$. - Point $B$ has coordinates $(0, \frac{7\pi}{12})$. Find slopes of tangent and normal at $A$ and $B$. At $A$: $$y = \frac{\pi}{4} \implies 3y + \frac{\pi}{4} = 3 \cdot \frac{\pi}{4} + \frac{\pi}{4} = \pi + \frac{\pi}{4} = \frac{5\pi}{4}.$$ Calculate $\frac{dy}{dx}$ at $A$: $$\frac{dy}{dx} = \frac{1}{2 \cos(3y + \frac{\pi}{4})} = \frac{1}{2 \cos(\frac{5\pi}{4})}.$$ Since $\cos(\frac{5\pi}{4}) = -\frac{\sqrt{2}}{2}$, $$\frac{dy}{dx}\bigg|_A = \frac{1}{2 \cdot (-\frac{\sqrt{2}}{2})} = \frac{1}{-\sqrt{2}} = -\frac{1}{\sqrt{2}}.$$ Slope of normal at $A$ is negative reciprocal: $$m_{normal,A} = \sqrt{2}.$$ At $B$: $$y = \frac{7\pi}{12} \implies 3y + \frac{\pi}{4} = 3 \cdot \frac{7\pi}{12} + \frac{\pi}{4} = \frac{21\pi}{12} + \frac{3\pi}{12} = 2\pi.$$ Calculate $\frac{dy}{dx}$ at $B$: $$\frac{dy}{dx} = \frac{1}{2 \cos(2\pi)} = \frac{1}{2 \cdot 1} = \frac{1}{2}.$$ Slope of tangent at $B$ is $\frac{1}{2}$. --- 5. **Find coordinates of $B$** Recall: $$x = \frac{2}{3} \sin(3y + \frac{\pi}{4}) = \frac{2}{3} \sin(2\pi) = 0.$$ So $B = (0, \frac{7\pi}{12})$. --- 6. **Find equation of normal at $A$** Point $A = (0, \frac{\pi}{4})$, slope $m = \sqrt{2}$. Equation: $$y - \frac{\pi}{4} = \sqrt{2}(x - 0) \implies y = \sqrt{2} x + \frac{\pi}{4}.$$ --- 7. **Find equation of tangent at $B$** Point $B = (0, \frac{7\pi}{12})$, slope $m = \frac{1}{2}$. Equation: $$y - \frac{7\pi}{12} = \frac{1}{2}(x - 0) \implies y = \frac{1}{2} x + \frac{7\pi}{12}.$$ --- 8. **Find intersection $D$ of normal at $A$ and tangent at $B$** Set equations equal: $$\sqrt{2} x + \frac{\pi}{4} = \frac{1}{2} x + \frac{7\pi}{12}.$$ Rearranged: $$\sqrt{2} x - \frac{1}{2} x = \frac{7\pi}{12} - \frac{\pi}{4}.$$ Calculate right side: $$\frac{7\pi}{12} - \frac{3\pi}{12} = \frac{4\pi}{12} = \frac{\pi}{3}.$$ Calculate left side: $$x(\sqrt{2} - \frac{1}{2}) = \frac{\pi}{3}.$$ Solve for $x$: $$x = \frac{\pi/3}{\sqrt{2} - \frac{1}{2}} = \frac{\pi/3}{\frac{2\sqrt{2} - 1}{2}} = \frac{\pi}{3} \cdot \frac{2}{2\sqrt{2} - 1} = \frac{2\pi}{3(2\sqrt{2} - 1)}.$$ Rationalize denominator: $$x = \frac{2\pi}{3(2\sqrt{2} - 1)} \cdot \frac{2\sqrt{2} + 1}{2\sqrt{2} + 1} = \frac{2\pi (2\sqrt{2} + 1)}{3[(2\sqrt{2})^2 - 1^2]} = \frac{2\pi (2\sqrt{2} + 1)}{3(8 - 1)} = \frac{2\pi (2\sqrt{2} + 1)}{21}.$$ --- **Final answers:** - (a) $y$-coordinates of $A$ and $B$: - $A: y = \frac{\pi}{4}$ - $B: y = \frac{7\pi}{12}$ - (b) $$\left(\frac{dy}{dx}\right)^2 = \frac{1}{4 - 9x^2}$$ with $p=4$, $q=9$. - (c) $x$-coordinate of $D$: $$x = \frac{2\pi (2\sqrt{2} + 1)}{21}.$$