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
 A: Let's take the origin of your coordinates at the vertex of your parabola.
(The minima of the parabola corresponds to its vertex.)
Since the parabola opens up, it's of the form
\begin{gather*}
y=ax^{2}\\
\end{gather*}
for any real number a.
From our figure, it seems that the abolute value of the x-coordinates of the endpoints of the parabola are the same, so let's take it as x.
The distance between the endpoints is 4.5.
So
\begin{gather*}
2x=4.5\\
x=\frac{9}{4}\\
\end{gather*}
The length of a curve is basically the integral
\begin{equation*}
\int ^{b}_{a}\sqrt{1+\left(\frac{dy}{dx}\right)^{2}} dx
\end{equation*}
From the equation of the parabola we took
\begin{equation*}
L=\int ^{\frac{9}{4}}_{-\frac{9}{4}}\sqrt{1+4a^{2} x^{2}} dx=6
\end{equation*}
The integral can be calculated by the method of integration by parts, and the resulting integrand turns out to be
\begin{equation*}
L=\frac{1}{2a}arcsinh\left(\frac{9a}{2}\right) +\frac{9}{8}\sqrt{81a^{2} +4}
\end{equation*}
Equate the length to 6 and you get the equation
\begin{equation*}
6=\frac{1}{2a}arcsinh\left(\frac{9a}{2}\right) +\frac{9}{8}\sqrt{81a^{2} +4}
\end{equation*}
This eqn is almost impossible to solve analytically.
However, since you just want an approximate answer, plugging this eqn into WolframAlpha gives the approximate solution as
\begin{equation*}
a=0.3553
\end{equation*}
As d is just the value of the function at its extreme endpoints,
\begin{equation*}
d=ax^{2} =0.3553\times \left(\frac{9}{4}\right)^{2} \approxeq 1.8
\end{equation*}
Hope this solves your queries:)
A: 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 $a=0.3553$.
