I need to design a second-order high-pass filter, gain 2 and cut frequency of 500 Hz. Using chebyshev response of 0.1 dB
Here's your diagram with some annotations:

Although I've never seen this particular arrangement before, I'll argue that the above can be reduced to:

Where \$R_2=\frac{R_a\,R_b}{R_a+R_b}\$ and \$A=\frac{1}{1+\frac{R_b}{R_a}}\$.
Since \$A\$ is in series with \$R_2\$, we can swap positions when performing the nodal analysis as seen below:
Using sympy and assuming \$C=C_1=C_2\$:
eq1 = Eq( va/ra + va/rb + va/(1/s/c) + va/(1/s/c), vp/(1/s/c) + vi/(1/s/c) + (A*vo)/r2 + 0/ra)
eq2 = Eq( vp/(1/s/c) + vp/r1, va/(1/s/c) + 0/r1 )
eq3 = Eq( vm/r3 + vm/r4, vo/r3 + 0/r4 )
eq4 = Eq( vo/r2 + vo/r3, vm/r3 + (va/A)/r2 + io )
eq5 = Eq( vp, vm )
ans = solve( [eq1, eq2, eq3, eq4, eq5], [io, va, vp, vm, vo] )
tf = simplify( ans[vo]/vi )
n0 = factor( -expand(fraction(simplify(ans[vo]/vi))[0]), s )
d0 = factor( -expand(fraction(simplify(ans[vo]/vi))[1]), s )
np = Poly( n0, s )
dp = Poly( d0, s )
w0 = sqrt( simplify(dp.coeffs()[2] / dp.coeffs()[0]) )
alpha = simplify( 1/2 * dp.coeffs()[1] / dp.coeffs()[0] )
zeta = simplify( alpha / w0 )
K = expand( simplify( np.coeffs()[0] / dp.coeffs()[0] ) )

So now I find:
$$\begin{align*}
R_2&amp;= R_a\mid\mid R_b = \frac{R_a\,R_b}{R_a+R_b}
\\\\
A&amp;=\frac1{1+\frac{R_b}{R_a}}
\\\\
K&amp;=1+\frac{R_3}{R_4}
\\\\
\omega_{_0}&amp;=\frac1{C\sqrt{R_1\,R_2}}
\\\\
\zeta&amp;=\sqrt{\frac{R_2}{R_1}}-\frac12\sqrt{\frac{R_1}{R_2}}\bigg[A\,K-1\bigg]
\\\\
\mathcal{H}\left(s\right)&amp;=K\frac{s^2}{s^2+2\zeta\,\omega_{_0}s+\omega_{_0}^{\:\!2}}
\end{align*}$$
The Chebyshev function you are looking to achieve (if I understand it) is:
eta0 = solve( Eq( 10**(-0.1/10), 1/(1+x**2) ), x )[1]
alp0 = arcsinh( 1 / eta1 ) / 2
chr0 = Poly( simplify((s-(-sinh(alp1)*sin(pi/4)+i*cosh(alp1)*cos(pi/4)))*(s-(-sinh(alp1)*sin(3*pi/4)+i*cosh(alp1)*cos(3*pi/4)))).n(), s )
w0chr = sqrt( chr0.coeffs()[2] / chr0.coeffs()[0] )
d0chr = chr0.coeffs()[1] / sqrt( chr0.coeffs()[0] * chr0.coeffs()[2] )
d0chr, w0chr

    (1.30317045127507, 1.82044969090790)

Let's look at a table from R. P. Sallen and E. L. Key's &quot;TR-50: A Practical Method of Designing RC Active Filters,&quot; 6 May 1954:

Similar values there.
As the frequency shape only depends on the value of \$d=2\,\zeta\$, we can set \$\zeta\approx 0.6516\$ for this case and then solve the above equations for the filter.
(Do keep in mind that I believe the \$0.1\:\text{dB}\$ down point will be \$\approx 1.82\times\$ your designed \$\omega_{_0}\$, based upon the above calculations I just performed. This could mean it's located at about \$f_c\approx 910 \:\text{Hz}\$. So this may not be expected or desired. If not, you'll have to take this factor into account, setting \$f_c= 500 \:\text{Hz}\$ and computing a now-reduced \$\omega_{_0}\$, before designing the filter. But I don't know what's expected. So I'm leaving it here.)
I think that spells out enough to get going. And I believe it is enough to say that your odd arrangement with \$R_a\$ and \$R_b\$ (which I've not seen) can be used. I'm not sufficiently familiar with it, yet, to know if there's a great purpose in arranging it with the extra resistor. Perhaps that will settle in, later. But at least I think it can be used to get where you need to go.