The equations of the directrices of the ellip | vedclass

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(A) Given equation: $9x^2 + 4y^2 - 18x - 16y - 11 = 0$ Rearranging terms: $9(x^2 - 2x) + 4(y^2 - 4y) = 11$ Completing the square: $9(x-1)^2 - 9 + 4(y-

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$y = 2 \pm \frac{9}{\sqrt{5}}$

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(A) Given equation: $9x^2 + 4y^2 - 18x - 16y - 11 = 0$ Rearranging terms: $9(x^2 - 2x) + 4(y^2 - 4y) = 11$ Completing the square: $9(x-1)^2 - 9 + 4(y-2)^2 - 16 = 11$ $9(x-1)^2 + 4(y-2)^2 = 36$ Dividing by $36$: $\frac{(x-1)^2}{4} + \frac{(y-2)^2}{9} = 1$ Here,$a^2 = 4$ and $b^2 = 9$. Since $b^2 > a^2$,the major axis is vertical. Eccentricity $e = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}$ The center is $(h, k) = (1, 2)$. The equations of the directrices for a vertical ellipse are $y - k = \pm \frac{b}{e}$. $y - 2 = \pm \frac{3}{\sqrt{5}/3} = \pm \frac{9}{\sqrt{5}}$ $y = 2 \pm \frac{9}{\sqrt{5}}$

The equations of the directrices of the ellipse $9x^2 + 4y^2 - 18x - 16y - 11 = 0$ are

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