(P) $\quad y=4 x^3-3 x \quad$ where $\cos \theta=x$ $ \begin{aligned} & \frac{d y}{d x}=12 x^2-3 \\ & \frac{d^2 y}{d x^2}+x \frac{d y}{d x}=\left(x^2-1\right) \cdot 24 x+x\left(12 x^2-3\right)=36 x^3-27 x=9\left(4 x^3-3 x\right)=9 y \end{aligned} $
Hence $\frac{1}{y}\left\{\left(x^2-1\right) \frac{d^2 y}{d x^2}+x \frac{d y}{d x}\right\}=9$
(Q) $\quad\left|\vec{a}_1 \times \vec{a}_2+\vec{a}_2 \times \vec{a}_3+\ldots+\vec{a}_{n-1} \times \vec{a}_n\right|=\left|\vec{a}_1 \cdot \vec{a}_2+\vec{a}_2 \cdot \vec{a}_3+\ldots+\vec{a}_{n-1} \cdot \vec{a}_n\right|$
Let $\left|\vec{a}_1\right|=\left|\vec{a}_2\right|=\ldots .=\left|\vec{a}_n\right|=\lambda$ (as centre is origin) More over angle between 2 consecutive $\vec{a}_i$ 's is $\frac{2 \pi}{n}$
Hence given equation reduces to
$ \begin{aligned} & (n-1) \lambda^2 \sin \left(\frac{2 \pi}{n}\right)=(n-1) \lambda^2 \cos \left(\frac{2 \pi}{n}\right) \\ \Rightarrow & \tan \left(\frac{2 \pi}{n}\right)=1 \quad \Rightarrow \quad \frac{2 \pi}{n}=\frac{\pi}{4} \quad \Rightarrow \quad n=8 \end{aligned} $
(R) Equation of normal $\frac{6 x}{h}-\frac{3 y}{1}=3$ (Equation of normal is $\frac{a^2 x}{x_1}-\frac{b^2 y}{y_1}=a^2-b^2$ )
$ \text { slope }=\frac{6}{3 h}=1 \text { (as it is perpendicular to } z+y=1 \text { ) } \quad \Rightarrow \quad h=2 $
$\left.\begin{array}{ll}\Rightarrow & \frac{1}{2 x+1}+\frac{1}{4 x+1}=\frac{2}{x^2}\end{array} \quad \Rightarrow \quad \frac{6 x+2}{8 x^2+6 x}=\frac{2}{x^2} ~(2 x+1)(4 x+1)\right) \quad \Rightarrow \quad 3 x^3-7 x^2-6 x=0 ~\left(-\frac{2}{3}\right.$ is rejected $)$