The length of the latus rectum of the parabol | vedclass

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(D) The given equation is $2\{(x - 1)^2 + (y - 3)^2\} = (x + y - 1)^2$.Dividing by $2$,we get $(x - 1)^2 + (y - 3)^2 = \frac{1}{2}(x + y - 1)^2$.This

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$3 \sqrt{2}$

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(D) The given equation is $2\{(x - 1)^2 + (y - 3)^2\} = (x + y - 1)^2$.Dividing by $2$,we get $(x - 1)^2 + (y - 3)^2 = \frac{1}{2}(x + y - 1)^2$.This is in the form $SP^2 = e^2 PM^2$,where $S(1, 3)$ is the focus and $x + y - 1 = 0$ is the directrix.Here,$e^2 = \frac{1}{2}$,so $e = \frac{1}{\sqrt{2}}$.The distance from the focus $S(1, 3)$ to the directrix $L: x + y - 1 = 0$ is $d = \frac{|1 + 3 - 1|}{\sqrt{1^2 + 1^2}} = \frac{3}{\sqrt{2}}$.The length of the latus rectum is $2e^2d$ or $2 	imes 	ext{distance from focus to directrix} 	imes e^2$ is not correct here; rather,for a parabola $SP = PM$,the definition is $SP^2 = PM^2$. Wait,the standard form is $SP^2 = PM^2$. Here,$SP^2 = \frac{1}{2} PM^2$. This implies $e^2 = 1/2$,which is an ellipse. However,the question states it is a parabola. Let's re-examine: $2\{(x-1)^2 + (y-3)^2\} = (x+y-1)^2 \implies (x-1)^2 + (y-3)^2 = \frac{(x+y-1)^2}{2} = \left(\frac{x+y-1}{\sqrt{2}}ight)^2$.This is $SP^2 = PM^2$ where $PM = \frac{|x+y-1|}{\sqrt{2}}$. This is a parabola with focus $S(1, 3)$ and directrix $x+y-1=0$.The distance $d$ from $S$ to the directrix is $\frac{|1+3-1|}{\sqrt{1^2+1^2}} = \frac{3}{\sqrt{2}}$.The length of the latus rectum of a parabola is $2d = 2 	imes \frac{3}{\sqrt{2}} = 3\sqrt{2}$.

The length of the latus rectum of the parabola $2\{(x - 1)^2 + (y - 3)^2\} = (x + y - 1)^2$ is:

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