If the line 2x - 3y + 4 = 0 cuts the ellipse | vedclass

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(B) The equation of the ellipse is $\frac{x^2}{9} + \frac{y^2}{25} = 1$. Given the line $2x - 3y + 4 = 0$,we can write $y = \frac{2x + 4}{3}$. Substit

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$4$

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(B) The equation of the ellipse is $\frac{x^2}{9} + \frac{y^2}{25} = 1$. Given the line $2x - 3y + 4 = 0$,we can write $y = \frac{2x + 4}{3}$. Substituting this into the ellipse equation: $25x^2 + 9(\frac{2x + 4}{3})^2 = 225$. $25x^2 + (2x + 4)^2 = 225$ $\Rightarrow 25x^2 + 4x^2 + 16x + 16 = 225$ $\Rightarrow 29x^2 + 16x - 209 = 0$. The midpoint $\alpha$ of the $x$-coordinates is $\frac{x_1 + x_2}{2} = \frac{-16/29}{2} = -\frac{8}{29}$. Similarly,for $y$,$x = \frac{3y - 4}{2}$. Substituting into the ellipse equation: $25(\frac{3y - 4}{2})^2 + 9y^2 = 225$. $\frac{25}{4}(9y^2 - 24y + 16) + 9y^2 = 225$ $\Rightarrow 225y^2 - 600y + 400 + 36y^2 = 900$ $\Rightarrow 261y^2 - 600y - 500 = 0$. The midpoint $\beta$ of the $y$-coordinates is $\frac{y_1 + y_2}{2} = \frac{600/261}{2} = \frac{300}{261} = \frac{100}{87}$. Now,$3\beta - 2\alpha = 3(\frac{100}{87}) - 2(-\frac{8}{29}) = \frac{100}{29} + \frac{16}{29} = \frac{116}{29} = 4$.

If the line $2x - 3y + 4 = 0$ cuts the ellipse $x = 3 \cos \theta, y = 5 \sin \theta$ at points $A$ and $B$,and $(\alpha, \beta)$ is the midpoint of $\overline{AB}$,then $3\beta - 2\alpha =$

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