1. **Problem statement:** Find the y-coordinates of points A and B where the curve $x=\frac{2}{3}\sin(3y+\frac{\pi}{4})$ intersects the y-axis, i.e., where $x=0$. 2. **Step for part (a):** Since $x=0$ at points A and B, set: $$0=\frac{2}{3}\sin\left(3y+\frac{\pi}{4}\right)$$ which implies $$\sin\left(3y+\frac{\pi}{4}\right)=0$$ 3. **Solve the trigonometric equation:** The general solution for $\sin \theta=0$ is: $$\theta = n\pi, \quad n \in \mathbb{Z}$$ So, $$3y+\frac{\pi}{4} = n\pi$$ 4. **Isolate $y$:** $$y = \frac{n\pi - \frac{\pi}{4}}{3} = \frac{n\pi}{3} - \frac{\pi}{12}$$ 5. **Apply domain restriction:** Given $\frac{\pi}{12} < y < \frac{3\pi}{4}$, find integer values of $n$ such that $y$ lies in this interval. - For $n=1$: $$y = \frac{\pi}{3} - \frac{\pi}{12} = \frac{4\pi}{12} - \frac{\pi}{12} = \frac{3\pi}{12} = \frac{\pi}{4}$$ - For $n=2$: $$y = \frac{2\pi}{3} - \frac{\pi}{12} = \frac{8\pi}{12} - \frac{\pi}{12} = \frac{7\pi}{12}$$ Both $\frac{\pi}{4}$ and $\frac{7\pi}{12}$ lie within the domain. 6. **Therefore,** - Point A has $y=\frac{\pi}{4}$ - Point B has $y=\frac{7\pi}{12}$ --- 7. **Part (b):** Given: $$x=\frac{2}{3}\sin\left(3y+\frac{\pi}{4}\right)$$ We want to find $\left(\frac{dy}{dx}\right)^2$ in the form: $$\left(\frac{dy}{dx}\right)^2 = \frac{1}{p - qx^2}$$ 8. **Find $\frac{dx}{dy}$:** $$\frac{dx}{dy} = \frac{2}{3} \cdot \cos\left(3y+\frac{\pi}{4}\right) \cdot 3 = 2 \cos\left(3y+\frac{\pi}{4}\right)$$ 9. **Find $\frac{dy}{dx}$:** $$\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} = \frac{1}{2 \cos\left(3y+\frac{\pi}{4}\right)}$$ 10. **Square both sides:** $$\left(\frac{dy}{dx}\right)^2 = \frac{1}{4 \cos^2\left(3y+\frac{\pi}{4}\right)}$$ 11. **Express $\cos^2$ in terms of $\sin^2$:** $$\cos^2 \theta = 1 - \sin^2 \theta$$ Using $\theta = 3y + \frac{\pi}{4}$, $$\cos^2\left(3y+\frac{\pi}{4}\right) = 1 - \sin^2\left(3y+\frac{\pi}{4}\right)$$ 12. **Recall from original equation:** $$x = \frac{2}{3} \sin\left(3y+\frac{\pi}{4}\right) \implies \sin\left(3y+\frac{\pi}{4}\right) = \frac{3x}{2}$$ 13. **Substitute:** $$\cos^2\left(3y+\frac{\pi}{4}\right) = 1 - \left(\frac{3x}{2}\right)^2 = 1 - \frac{9x^2}{4}$$ 14. **Therefore:** $$\left(\frac{dy}{dx}\right)^2 = \frac{1}{4 \left(1 - \frac{9x^2}{4}\right)} = \frac{1}{4 - 9x^2}$$ 15. **Identify $p$ and $q$:** $$p = 4, \quad q = 9$$ --- 16. **Part (c):** Find the x-coordinate of point D, the intersection of the normal to C at A and the tangent to C at B. 17. **Find slope at A:** At $y=\frac{\pi}{4}$, from step 12: $$\sin\left(3\cdot \frac{\pi}{4} + \frac{\pi}{4}\right) = \sin\left(\frac{3\pi}{4} + \frac{\pi}{4}\right) = \sin(\pi) = 0$$ From step 9: $$\frac{dy}{dx} = \frac{1}{2 \cos(\pi)} = \frac{1}{2 \cdot (-1)} = -\frac{1}{2}$$ 18. **Slope of normal at A:** $$m_{normal} = -\frac{1}{m_{tangent}} = -\frac{1}{-\frac{1}{2}} = 2$$ 19. **Find slope at B:** At $y=\frac{7\pi}{12}$, $$3y + \frac{\pi}{4} = 3 \cdot \frac{7\pi}{12} + \frac{\pi}{4} = \frac{21\pi}{12} + \frac{3\pi}{12} = 2\pi$$ $$\cos(2\pi) = 1$$ From step 9: $$\frac{dy}{dx} = \frac{1}{2 \cdot 1} = \frac{1}{2}$$ 20. **Equations of lines:** - Normal at A passes through $(0, \frac{\pi}{4})$ with slope 2: $$y - \frac{\pi}{4} = 2(x - 0) \implies y = 2x + \frac{\pi}{4}$$ - Tangent at B passes through $(0, \frac{7\pi}{12})$ with slope $\frac{1}{2}$: $$y - \frac{7\pi}{12} = \frac{1}{2}(x - 0) \implies y = \frac{1}{2}x + \frac{7\pi}{12}$$ 21. **Find intersection D:** Set equal: $$2x + \frac{\pi}{4} = \frac{1}{2}x + \frac{7\pi}{12}$$ 22. **Solve for $x$:** $$2x - \frac{1}{2}x = \frac{7\pi}{12} - \frac{\pi}{4}$$ $$\frac{3}{2}x = \frac{7\pi}{12} - \frac{3\pi}{12} = \frac{4\pi}{12} = \frac{\pi}{3}$$ $$x = \frac{2}{3} \cdot \frac{\pi}{3} = \frac{2\pi}{9}$$ **Final answers:** - (a) $y_A = \frac{\pi}{4}$, $y_B = \frac{7\pi}{12}$ - (b) $\left(\frac{dy}{dx}\right)^2 = \frac{1}{4 - 9x^2}$ with $p=4$, $q=9$ - (c) $x_D = \frac{2\pi}{9}$