The length of the latusrectum of the parabola $20\left(x^2+y^2-6 x-2 y+10\right)=(4 x-2 y-5)^2$, is

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Given, equation of parabola is $\begin{aligned} 20\left(x^2+y^2-6 x-2 y+10\right) & =(4 x-2 y-5)^2 \ \Rightarrow \quad\left(x^2+y^2-6 x-2 y+10\right) & =\left(\frac{4 x-2 y-5}{\sqrt{20}}\right)^2 \ \Rightarrow \quad(x-3)^2+(y-1)^2 & =\left(\frac{4 x-2 y-5}{\sqrt{20}}\right)^2 \quad \ldots (i) \end{aligned}$ In Eq. (i), focus is $(3,1)$ and equation of directrix is $4 x-2 y-5=0$. So, distance from focus to directrix is $\frac{|12-2-5|}{\sqrt{20}}$ $=\frac{\sqrt{5}}{2}=2 a$ Now, length of latursrectum $=4 a$ $=2(2 a)=2 \frac{\sqrt{5}}{2}=\sqrt{5}$

The length of the latusrectum of the parabola \(20\left(x^2+y^2-6 x-2 y+10\right)=(4 x-2 y-5)^2\), is

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