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\displaystyle \text{Newton's second law on }P_2:\; m_2g-T=m_2a \\[br]\text{Newton's second law on }P_1:\; T-m_1g=m_1a\\[br]\text{Dividing, }\frac{m_2g-T}{T-m_1g} = \frac{m_2}{m_1}\\[br]m_1m_2g - m_1T = m_2T - m_1m_2g\\[br]T(m_1+m_2) = 2m_1m_2g\\[br]T = \frac{2m_1m_2g}{m_1+m_2}.\\[br]P_3 \text{ remains at rest if there is no acceleration} .\\[br]\Leftrightarrow m_3g = 2T.\\[br]m_3g = \frac{4m_1m_2g}{m_1+m_2}[br]\text{and result follows} .
\displaystyle r = \frac{\l}{1+e\cos\theta},\; \phi = 45^\text{o} + \theta
\displaystyle \therefore r = \frac{\l}{1+e\cos (\phi -45^\text{o} )} = \frac{l}{\sqrt{2}} \cdot \frac{1}{(1/\sqrt{2}) + e\cos\phi + e\sin\phi}.\\[br]\tfrac{l}{\sqrt{2}} \cdot x = \frac{\cos\phi}{(1/\sqrt{2}) + e\cos\phi + e\sin\phi} \\[br]\tfrac{l}{\sqrt{2}} \dot{x} = \frac{-[(1/\sqrt{2}) + e\cos\phi + e\sin\phi ]\sin\phi - \cos\phi (e\cos\phi - e\sin\phi)}{[(1/\sqrt{2}) + e\cos\phi + e\sin\phi]^2} \\[br]=\frac{-\tfrac{1}{\sqrt{2}} \sin\phi - e}{[(1/\sqrt{2}) + e\cos\phi + e\sin\phi]^2} . \\[br]\tfrac{l}{\sqrt{2}} \cdot y = \frac{\sin\phi}{(1/\sqrt{2}) + e\cos\phi + e\sin\phi} \\[br]\tfrac{l}{\sqrt{2}} \dot{y} = \frac{[(1/\sqrt{2}) + e\cos\phi + e\sin\phi ]\cos\phi - \sin\phi (e\cos\phi - e\sin\phi)}{[(1/\sqrt{2}) + e\cos\phi + e\sin\phi]^2} \\[br]=\frac{\tfrac{1}{\sqrt{2}} \cos\phi + e}{[(1/\sqrt{2}) + e\cos\phi + e\sin\phi]^2} . \\[br]\therefore \frac{\text{d}y}{\text{d}x} = \dot{y} / \dot{x} = \frac{\cos\phi + \sqrt{2} e}{-(\sin\phi + \sqrt{2} e)} .\\[br]\theta = \alpha \Rightarrow \phi = 45^\text{o} + \alpha \\[br]\therefore \bigg( \frac{\text{d}y}{\text{d}x} \bigg)_{\theta =\alpha} = \frac{\cos(45^\text{o} + \alpha) + \sqrt{2} e}{-(\sin(45^\text{o} + \alpha) + \sqrt{2} e)} \\[br]= \frac{\sqrt{2} \cos\alpha - \sqrt{2} \sin\alpha + \sqrt{2} e}{-(\sqrt{2} \sin\alpha + \sqrt{2} \cos\alpha + \sqrt{2} e)} = \frac{\sin\alpha - \cos\alpha - e}{\sin\alpha + \cos\alpha + e}.
\displaystyle x = OP\cos\phi + PB\cos (\phi +\theta ) = 3\cos\phi + \cos (-3\phi ) = 3\cos\phi + \cos 3\phi \\[br]y = OP\sin\phi + PB\sin (\phi +\theta ) = 3\sin\phi + \sin (-3\phi ) = 3\sin\phi - \sin 3\phi \\[br]\therefore (x, y) = (3\cos\phi + \cos 3\phi , 3\sin\phi - \sin 3\phi ).\\[br]A = \int_0^{2\pi} x \frac{\text{d}y}{\text{d}\phi} \text{d}\phi = \int_0^{2\pi} (3\cos\phi + \cos 3\phi )(3\cos\phi + 3\cos 3\phi ) \text{d}\phi \\[br]= \int_0^{2\pi} (9\cos^2 \phi - 6\cos\phi\cos 3\phi - 3\cos^2 3\phi) \text{d}\phi \\[br]= \frac{1}{2} \int_0^{2\pi} (9\cos 2\phi + 9 - 6\cos 2\phi - 6\cos 4\phi - 3\cos 6\phi - 3) \text{d}\phi \\[br]= \frac{1}{2} \left[6\phi + \tfrac{9}{2} \sin 2\phi - 3\sin 2\phi - \tfrac{3}{4} \sin 4\phi - \tfrac{1}{2} \sin 6\phi \right]_0^{2\pi} \\[br]= 3\left[\phi\right]_0^{2\pi} = 6\pi .
\displaystyle[br]\text{Let } \overrightarrow{AB} = \mathbf{a}, \overrightarrow{AD} = \mathbf{b} .\\[br]\overrightarrow{AP} = \lambda\mathbf{a} \\[br]\overrightarrow{AR} = \mu\mathbf{b}\\[br]\overrightarrow{AQ} = \lambda\mathbf{a} + \mu\mathbf{b}\\[br]\overrightarrow{AC} = \mathbf{a} + \mathbf{b}\\[br]\overrightarrow{CQ} = (\lambda -1)\mathbf{a} + (\mu -1)\mathbf{b}\[br]\overrightarrow{CX} = k_0 [(\lambda -1)\mathbf{a} + (\mu -1)\mathbf{b} ]\\[br]\therefore \overrightarrow{AX} = (1 + k_0 \lambda - k_0)\mathbf{a} + (1 + k_0\mu - k_0)\mathbf{b} .
\overrightarrow{AB} = \mathbf{a}\\[br]\overrightarrow{RB} = \mathbf{a} - \mu\mathbf{b}\\[br]\overrightarrow{RX} = k_1 [\mathbf{a} - \mu\mathbf{b} ]\\[br]\overrightarrow{AX} = k_1\mathbf{a} + \mu (1-k_1)\mathbf{b} .
\therefore k_1\mathbf{a} + \mu (1-k_1)\mathbf{b} = (1 + k_0 \lambda - k_0)\mathbf{a} + (1 + k_0\mu - k_0)\mathbf{b} \\[br]\text{Equating coefficients, }\\[br]1 + k_0 \lambda - k_0 = k_1 \Rightarrow k_0 = \frac{k_1 -1}{\lambda -1} \\[br]1 + k_0\mu - k_0 = \mu (1-k_1) \Rightarrow k_0 = \frac{\mu (1-k_1)-1}{\mu -1} \\[br]\therefore \frac{k_1 -1}{\lambda -1} = \frac{\mu (1-k_1)-1}{\mu -1} \\[br](k_1 -1)(\mu -1) = (\mu (1-k_1)-1)(\lambda -1) \Rightarrow k_1 = \frac{\lambda (1-\mu )}{1-\lambda\mu} .
\overrightarrow{DP} = \lambda\mathbf{a} - \mathbf{b} \\[br]\overrightarrow{DY} = k_2(\lambda\mathbf{a} - \mathbf{b} )\\[br]\overrightarrow{AY} = k_2\lambda\mathbf{a} + (1-k_2)\mathbf{b} .
\overrightarrow{RY} = k_3[\mathbf{a} -\mu\mathbf{b} ]\\[br]\overrightarrow{AY} = k_3\mathbf{a} + \mu 91-k_3)\mathbf{b} .
\text{Equating coefficients and solving simultaneously gives} \\[br]k_3 = \frac{\lambda (1-\mu )}{1-\lambda\mu} = k_1. \\[br]\therefore \overrightarrow{RX} = \overrightarrow{RY} \Rightarrow X \text{ and } Y \text{ are the same point} .\\[br]\Rightarrow Z \text{ must also be this point} .
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