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The length of the latusrectum of the conic $25\left[(x-2)^2+(y-3)^2\right]=(3 x-4 y+7)^2$ is
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The correct answer is:
$\frac{2}{5}$
Given, 25[(x-2) $\left.)^2+(y-3)^2\right]=[3 x-4 y+7]^2$ is
$$
\Rightarrow(x-2)^2+(y-3)^2=\left(\frac{3 x-4 y+7}{\sqrt{25}}\right)^2
$$
Its focus $\equiv(2,3)$, Directrix $\rightarrow 3 x-4 y+7$
$\therefore L=$ length of latus rectum $=2$ (distance between focus and directrix)
$$
=2\left|\frac{3 \times 2-4 \times 3+7}{\sqrt{3^2+4^2}}\right|=2\left|\frac{6-12+7}{5}\right|=\frac{2}{5}
$$
$$
\Rightarrow(x-2)^2+(y-3)^2=\left(\frac{3 x-4 y+7}{\sqrt{25}}\right)^2
$$
Its focus $\equiv(2,3)$, Directrix $\rightarrow 3 x-4 y+7$
$\therefore L=$ length of latus rectum $=2$ (distance between focus and directrix)
$$
=2\left|\frac{3 \times 2-4 \times 3+7}{\sqrt{3^2+4^2}}\right|=2\left|\frac{6-12+7}{5}\right|=\frac{2}{5}
$$
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