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If $m$ is the length of the latus rectum and $n$ is the length of the major axis of the ellipse $25 x^2+16 y^2-150 x-64 y-111=0$, then the ordered pair $(m, n)=$
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The correct answer is:
$\left(\frac{32}{5}, 10\right)$
Given equation of ellipse is $25 \mathrm{x}^2+16 \mathrm{y}^2-150 \mathrm{x}-$
$64 \mathrm{y}-111=0$
On arranging it we get,
$$
\begin{aligned}
& \frac{(\mathrm{x}-3)^2}{4^2}+\frac{(\mathrm{y}-2)^2}{5^2}=1 \\
& \Rightarrow \mathrm{a}=4, \text { and } \mathrm{b}=5 \\
& \text { now } \eta=\frac{2 \mathrm{a}^2}{\mathrm{~b}}=\frac{2 \times 16}{5}=\frac{32}{5} \\
& \mathrm{~m}=2 \mathrm{~b}=2 \times 5=10 \\
& \text { Hence }(\mathrm{n}, \mathrm{m}) \equiv\left(\frac{32}{5}, 10\right)
\end{aligned}
$$
$64 \mathrm{y}-111=0$
On arranging it we get,
$$
\begin{aligned}
& \frac{(\mathrm{x}-3)^2}{4^2}+\frac{(\mathrm{y}-2)^2}{5^2}=1 \\
& \Rightarrow \mathrm{a}=4, \text { and } \mathrm{b}=5 \\
& \text { now } \eta=\frac{2 \mathrm{a}^2}{\mathrm{~b}}=\frac{2 \times 16}{5}=\frac{32}{5} \\
& \mathrm{~m}=2 \mathrm{~b}=2 \times 5=10 \\
& \text { Hence }(\mathrm{n}, \mathrm{m}) \equiv\left(\frac{32}{5}, 10\right)
\end{aligned}
$$
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