The length of the latus rectum of the parabol | vedclass

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(C) The given equation of the parabola is $20(x^2+y^2-6x-2y+10) = (4x-2y-5)^2$. Dividing by $20$,we get $(x-3)^2 + (y-1)^2 = \left(\frac{4x-2y-5}{\sqr

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$\sqrt{5}$

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(C) The given equation of the parabola is $20(x^2+y^2-6x-2y+10) = (4x-2y-5)^2$. Dividing by $20$,we get $(x-3)^2 + (y-1)^2 = \left(\frac{4x-2y-5}{\sqrt{20}}ight)^2$. This is in the form $SP^2 = PM^2$,where $S(3,1)$ is the focus and $4x-2y-5=0$ is the directrix. The distance from the focus to the directrix is $2a = \frac{|4(3)-2(1)-5|}{\sqrt{4^2+(-2)^2}} = \frac{|12-2-5|}{\sqrt{16+4}} = \frac{5}{\sqrt{20}} = \frac{5}{2\sqrt{5}} = \frac{\sqrt{5}}{2}$. The length of the latus rectum is $4a = 2(2a) = 2 	imes \frac{\sqrt{5}}{2} = \sqrt{5}$.

The length of the latus rectum of the parabola $20(x^2+y^2-6x-2y+10) = (4x-2y-5)^2$ is

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