The length of the latus rectum of the conic 2 | vedclass

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(D) The given equation is $25[(x-2)^2+(y-3)^2]=(3x-4y+7)^2$. Dividing by $25$,we get $(x-2)^2+(y-3)^2 = \left(\frac{3x-4y+7}{5}\right)^2$. This is in

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$\frac{4}{5}$

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(D) The given equation is $25[(x-2)^2+(y-3)^2]=(3x-4y+7)^2$. Dividing by $25$,we get $(x-2)^2+(y-3)^2 = \left(\frac{3x-4y+7}{5}ight)^2$. This is in the form $SP^2 = PM^2$,where $S(2,3)$ is the focus and $3x-4y+7=0$ is the directrix. Here,the distance from the focus to the directrix is $a = \left|\frac{3(2)-4(3)+7}{\sqrt{3^2+(-4)^2}}ight| = \left|\frac{6-12+7}{5}ight| = \frac{1}{5}$. The length of the latus rectum of a parabola is $4a$. Therefore,length $= 4 	imes \frac{1}{5} = \frac{4}{5}$.

The length of the latus rectum of the conic $25[(x-2)^2+(y-3)^2]=(3x-4y+7)^2$ is

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