How to find an ellipse equation with non-symmetrical foci values? I have tried a lot but i do not seem to find an equation.
Question is the following:
Find an equation of the specified curve, that is, of the locus of points the sum of whose distances from the points $(2,3)$ and $(4,1)$ is $8$.
I first tried to translate and rotate the axis to see the ellipse like its center lies on the origin $(0,0)$. It seemed to be tough. I know I can use the general equation with center $(h,k)$ but then there will not be any rotation though the major axis is rotated.

Let the distances to the foci be $d_1,d_2$ and their sum be $d$. We have
$$d=d_1+d_2$$
Squaring,
$$d^2-d_1^2-d_2^2=2d_1d_2.$$
Squaring a second time,
$$d^4-2d^2(d_1^2+d_2^2)+(d_1^2+d_2^2)^2=4d_1^2d_2^2,$$
or
$$d^4-2d^2(d_1^2+d_2^2)+(d_1^2-d_2^2)^2=0.$$
Expanding the distances, we have
$$d_1^2+d_2^2=2x^2+2y^2-2(x_1+x_2)x-2(y_1+y_2)y+x_1^2+x_2^2+y_1^2+y_2^2\\
=2x^2+2y^2-2\sigma_xx-2\sigma_yy+\Sigma$$
and
$$d_1^2-d_2^2=-2(x_1-x_2)x-2(y_1-y_2)y+x_1^2-x_2^2+y_1^2-y_2^2\\
=-2\delta_xx-2\delta_yy+\Delta,$$
and
$$(d_1^2-d_2^2)^2=4\delta_x^2x^2+8\delta_x\delta_yxy+4\delta_y^2y^2-4\delta_x\Delta x-4\delta_y\Delta y+\Delta^2.$$
Finally, we get the conic
$$4(\delta_x^2-d^2)x^2+8\delta_x\delta_yxy+4(\delta_y^2-d^2)y^2-4(\delta_x\Delta+d^2\sigma_x) x-4(\delta_y\Delta+d^2\sigma_y)y\\
+\Delta^2-2d^2\Sigma+d^4=0.$$

Check:
With the usual placement of the foci, $(f,0)$ and $(-f,0)$, we have $\sigma_x=\sigma_y=\delta_y=\Delta=0$, $\delta_x=2f$ and $\Sigma=2f^2$. The equation reduces to
$$4(d^2-4f^2)x^2+4d^2y^2=d^2(d^2-4f^2).$$
