How is the depth of a parabolic mirror calculated How is the depth of a parabolic mirror calculated. I couldn't find a similar question/answer.
I have this specific problem to solve (not related to parabolic mirrors, but is a starting point, and not interested in a focal point):

*

*L=6"

*D=4.5"

*What is d?


Thanks for your help

Staring from @Ishraaq Parvez's answer, as said, the equation
$$L=\frac{1}{2 a}\sinh ^{-1}\left(\frac{9 }{2}a\right)+\frac{9}{8} \sqrt{81 a^2+4}$$ requires some numerical method. However we can make nice approximations.
Let $x=\frac{9 }{2}a$ and $k=\frac{4 }{9}L$ to obtain
$$k=\sqrt{x^2+1}+\frac{\sinh ^{-1}(x)}{x}$$ and expand the rhs as a Taylor series
$$k=2+\frac{x^2}{3}-\frac{x^4}{20}+\frac{x^6}{56}-\frac{5 x^8}{576}+\frac{7
   x^{10}}{1408}+O\left(x^{12}\right)$$ Now, let $y=x^2$ and use series reversion to obtain
$$y=t+\frac{3 t^2}{20}-\frac{3 t^3}{350}+\frac{23 t^4}{8400}-\frac{5889
   t^5}{5390000}+O\left(t^6\right)$$ where $t=3(k-2)$.
Using $L=6$, $k=\frac{8}{3}$ then $t=2$, this would give
$$y\sim \frac{2567266}{1010625}\implies x\sim \frac 1 {175}\sqrt{\frac{2567266}{33}}\implies a\sim \frac 2 {1575}\sqrt{\frac{2567266}{33}}$$ Converted to decimals, this gives
$a=0.3542$ while the solution is $a=0.3553$.
