The distance of a real object from the focus of a convex mirror of radius of curvature ‘a’ is ‘b’. Then the distance of the image from the focus is

The focal length of convex mirror is f = a/2

The real object distance u = b - a/2

$$\begin{gathered} \frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}} \\ \frac{1}{\mathrm{v}}-\frac{1}{\mathrm{b}-\left(\mathrm{a}/2\right)}=\frac{2}{\mathrm{a}} \end{gathered}$$

solving we get

$\mathrm{v}=\frac{\mathrm{a}\left(2\mathrm{b}-\mathrm{a}\right)}{4\mathrm{b}}$

distance of image from focus is

$\frac{\mathrm{a}}{2}-\mathrm{v}=\frac{{\mathrm{a}}^2}{4\mathrm{b}}$