Step 1:

Let $V(0,0)$ be the centre of the parabolic mirror whose focus is $8$cm from $V$ (i.e) at $(8,0)$

$VF$ is the axis of the mirror,and it bisects the diameter $AB$ of the mirror at a distance of $25$cm from the focus.

Let $A$ be the point $(25,y)$ which lies on the parabola of which the mirror is a part.

Step 2:

Here $a=8$

Therefore the equation of the parabola is $y^2=4ax$

$\Rightarrow 4\times 8x$

$y^2=32x$

Step 3:

Substituting the coordinates $(25,y)$

$y^2=32\times 25$

$y=\pm \sqrt{32\times 25}$

$\Rightarrow y=\pm 20\sqrt 2$

The points $A$ and $B$ are $(25,20\sqrt 2),(25,-20\sqrt 2)$

The diameter is $AB=40\sqrt 2$cm.