Find the length of Latusrectum of the ellipse $(10x-5)^2+(10y-5)^2=(3x+4y-1)^2$

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Converting this into the standard form $$(x-\frac 12)^2+(y-\frac 12)^2=\frac 14 \left(\frac{3x+4y-1}{5}\right)^2$$

$$\sqrt {\left (x-\frac 12 \right )^2+\left (y-\frac 12 \right )^2}=\frac 12 \left (\frac{3x+4y-1}{5} \right )$$

So the focus is $(\frac 12, \frac 12)$ and eccentricity is $\frac 12$

The distance of focus from directrix is $a(\frac 1e-e)$

So $$a(2-\frac 12) =\frac {1.5+2-1}{5}$$ $$a\frac 32 =\frac12$$ $$a=\frac 13$$

And $$b^2=\frac 19 (\frac 34)$$ $$b^2=\frac {1}{12}$$

Then $$LR=\frac {2b^2}{a}=\frac{1}{18}$$

I think I am making a coneptual mistake in the equation of ellipse, but I can’t pinpoint it.

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Lastly $$LR=\frac{2b^2}{a}=\frac{2/12}{1/3}=\frac{1}{2}.$$