If m is the length of the latus rectum and n | vedclass

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(B) The given equation of the ellipse is $25x^2+16y^2-150x-64y-111=0$. Rearranging the terms: $25(x^2-6x) + 16(y^2-4y) = 111$ $25(x-3)^2 - 225 + 16(y-

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$\left(\frac{32}{5}, 10\right)$

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(B) The given equation of the ellipse is $25x^2+16y^2-150x-64y-111=0$. Rearranging the terms: $25(x^2-6x) + 16(y^2-4y) = 111$ $25(x-3)^2 - 225 + 16(y-2)^2 - 64 = 111$ $25(x-3)^2 + 16(y-2)^2 = 400$ Dividing by $400$: $\frac{(x-3)^2}{16} + \frac{(y-2)^2}{25} = 1$ Here,$a^2 = 16$ and $b^2 = 25$,so $a = 4$ and $b = 5$. Since $b > a$,the major axis is vertical. The length of the latus rectum $m = \frac{2a^2}{b} = \frac{2 	imes 16}{5} = \frac{32}{5}$. The length of the major axis $n = 2b = 2 	imes 5 = 10$. Thus,the ordered pair $(m, n) = \left(\frac{32}{5}, 10ight)$.

If $m$ is the length of the latus rectum and $n$ is the length of the major axis of the ellipse $25x^2+16y^2-150x-64y-111=0$,then the ordered pair $(m, n) =$

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