Let the line 2x + 3y - k = 0, k 0,intersect | vedclass

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(B) The center of the circle $x^2 + y^2 - 3x - 2y = 0$ is $(\frac{3}{2}, 1)$.Since the line segment $AB$ is the diameter of the circle,the center of t

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(B) The center of the circle $x^2 + y^2 - 3x - 2y = 0$ is $(\frac{3}{2}, 1)$.Since the line segment $AB$ is the diameter of the circle,the center of the circle must lie on the line $2x + 3y - k = 0$.Substituting the center $(\frac{3}{2}, 1)$ into the line equation: $2(\frac{3}{2}) + 3(1) - k = 0 \implies 3 + 3 - k = 0 \implies k = 6$.The equation of the ellipse is $x^2 + 9y^2 = 6^2 = 36$,which can be written as $\frac{x^2}{6^2} + \frac{y^2}{2^2} = 1$.Here,$a^2 = 36$ and $b^2 = 4$,so $a = 6$ and $b = 2$.The length of the latus rectum is $\frac{2b^2}{a} = \frac{2(4)}{6} = \frac{8}{6} = \frac{4}{3}$.Given $\frac{m}{n} = \frac{4}{3}$,where $m = 4$ and $n = 3$ are coprime.Therefore,$2m + n = 2(4) + 3 = 8 + 3 = 11$.

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