If the eccentricity of the ellipse frac{x^2}{ | vedclass

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(D) The foci of the ellipse are $(\pm ae, 0)$. Since the line $3x + 2y - 6 = 0$ is a focal chord, it must pass through one of the foci $(\pm ae, 0)$.G

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$14$

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(D) The foci of the ellipse are $(\pm ae, 0)$. Since the line $3x + 2y - 6 = 0$ is a focal chord, it must pass through one of the foci $(\pm ae, 0)$.Given $e = 2/3$, the foci are $(\pm \frac{2a}{3}, 0)$.Case $1$: Line passes through $(\frac{2a}{3}, 0) \implies 3(\frac{2a}{3}) + 2(0) - 6 = 0 \implies 2a = 6 \implies a = 3$.Case $2$: Line passes through $(-\frac{2a}{3}, 0) \implies 3(-\frac{2a}{3}) + 2(0) - 6 = 0 \implies -2a = 6 \implies a = -3$ (not possible as $a > 0$).Using $b^2 = a^2(1 - e^2)$, we get $b^2 = 3^2(1 - (2/3)^2) = 9(1 - 4/9) = 9(5/9) = 5$.Thus, $a^2 + b^2 = 3^2 + 5 = 9 + 5 = 14$.

If the eccentricity of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, (a > b)$ is $e = 2/3$ and its focal chord is $3x + 2y - 6 = 0$, then the value of $a^2 + b^2$ is...

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