The length of the latus rectum of the parabol | vedclass

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(C) The given equation of the parabola is $169\{(x-1)^2+(y-3)^2\}=(5x-12y+17)^2$. Dividing by $169$,we get: $(x-1)^2+(y-3)^2 = \left(\frac{5x-12y+17}{

Vedclass · Accepted Answer

$\frac{28}{13}$

Vedclass · Answer

(C) The given equation of the parabola is $169\{(x-1)^2+(y-3)^2\}=(5x-12y+17)^2$. Dividing by $169$,we get: $(x-1)^2+(y-3)^2 = \left(\frac{5x-12y+17}{13}ight)^2$. This is in the form $SP^2 = PM^2$,where $S$ is the focus $(1, 3)$ and $PM$ is the perpendicular distance from point $P(x, y)$ to the directrix $5x-12y+17=0$. The distance between the focus and the directrix is $2a$. $2a = \left|\frac{5(1)-12(3)+17}{\sqrt{5^2+(-12)^2}}ight| = \left|\frac{5-36+17}{13}ight| = \left|\frac{-14}{13}ight| = \frac{14}{13}$. The length of the latus rectum is $4a$. Since $2a = \frac{14}{13}$,then $4a = 2 	imes \frac{14}{13} = \frac{28}{13}$.

The length of the latus rectum of the parabola $169\{(x-1)^2+(y-3)^2\}=(5x-12y+17)^2$ is

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