The length of the latus rectum of the parabol | vedclass

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(B) The given equation of the parabola is $(x-2)^2+(y-3)^2=\frac{1}{25}(3x-4y+7)^2$.This is of the form $SP^2 = e^2 PM^2$,where $S$ is the focus,$P$ i

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$\frac{2}{5}$

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(B) The given equation of the parabola is $(x-2)^2+(y-3)^2=\frac{1}{25}(3x-4y+7)^2$.This is of the form $SP^2 = e^2 PM^2$,where $S$ is the focus,$P$ is a point $(x, y)$ on the parabola,$e$ is the eccentricity,and $PM$ is the perpendicular distance from $P$ to the directrix.Here,$e^2 = \frac{1}{25}$,so $e = \frac{1}{5}$.The focus $S$ is $(2, 3)$ and the directrix is $3x-4y+7=0$.The distance $d$ from the focus to the directrix is $d = \frac{|3(2)-4(3)+7|}{\sqrt{3^2+(-4)^2}} = \frac{|6-12+7|}{5} = \frac{1}{5}$.The length of the latus rectum is $2e 	imes d = 2 	imes \frac{1}{5} 	imes \frac{1}{5} = \frac{2}{25}$.Wait,re-evaluating the standard form: The equation is $(x-h)^2+(y-k)^2 = e^2 \frac{(ax+by+c)^2}{a^2+b^2}$.Here,$e^2 = \frac{1}{25} 	imes (3^2+(-4)^2) = \frac{25}{25} = 1$. Thus,it is a parabola.The distance from focus $(2, 3)$ to line $3x-4y+7=0$ is $d = \frac{|6-12+7|}{5} = \frac{1}{5}$.The length of the latus rectum is $2d = 2 	imes \frac{1}{5} = \frac{2}{5}$.

The length of the latus rectum of the parabola $(x-2)^2+(y-3)^2=\frac{1}{25}(3x-4y+7)^2$ is

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