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Q.
The length of the latus rectum of the parabola $25[(x-2) \left.^{2}+(y-4)^{2}\right]=(4 x-3 y+12)^{2}$ is
Conic Sections
Solution:
The given equation of the parabola can be written as
$(x-2)^{2}+(y-4)^{2}=\left(\frac{4 x-3 y+12}{\sqrt{(4)^{2}+(-3)^{2}}}\right)^{2}$
$\therefore $ The coordinates of focus are $(2,4)$ and the equation of directrix is $4 x-3 y+12=0$.
The distance of the focus from the directrix
$=\frac{|4(2)-3(4)+12|}{\sqrt{4^{2}+(-3)^{2}}}=\frac{8}{5}$
$\therefore $ The length of latus rectum $=2 \times \frac{8}{5}=\frac{16}{5}$.