A point source S is moving with a speed of $10 \mathrm{m}{\sec }^{-1}$ in $x-y$ plane as shown. The radius of curvature of a concave mirror is 4 m. What will be velocity vector of the image formed by paraxial rays

$$\begin{gathered} V_x=x component of velocity of the object=8m/s \\ {V}_{y }= y component of velocity of the object=6m/s \\ Now \frac{1}{v}+\frac{1}{u}=\frac{1}{f}\Rightarrow v=\frac{uf}{u-f}=\frac{\left(-6\right)\left(-2\right)}{-6+2}=-3m \\ V{\text{'}}_x=x component of velocity of image=\frac{dv}{dt}=-{\left(\frac{v}{u}\right)}^2\frac{du}{dt}=-{\left(\frac{-3}{-6}\right)}^2\left(-8\right)m/s \\ \Rightarrow V{\text{'}}_x=-2m/s \\ Now magnification m=-\frac{v}{u}\Rightarrow \frac{dm}{dt}=\frac{v{\displaystyle \frac{du}{dt}}-u{\displaystyle \frac{dv}{dt}}}{u^2}=\frac{\left(-3\right)8-\left(-6\right)\left(-2\right)}{36}=1 \\ Also \frac{h\text{'}}{h}=m \Rightarrow m=-\frac{v}{u}=-\frac{-3}{-6}=-\frac{1}{2} \\ \Rightarrow V{\text{'}}_y=\frac{dh\text{'}}{dt}=m\frac{dh}{dt}+h\frac{dm}{dt}=\left(-\frac{1}{2}\right)6+\left(1\right)\left(-1\right)=-4m/s \\ Therefore velocity of the image \overrightarrow{V\text{'}}=-2\widehat{i }-4\hat{j} m/s \end{gathered}$$