The equations of the latus rectum of the elli | vedclass

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(B) The given equation of the ellipse is $9x^2+4y^2-18x-8y-23=0$. Completing the square,we get: $9(x^2-2x+1) + 4(y^2-2y+1) = 23+9+4$ $9(x-1)^2 + 4(y-1

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$y=1 \pm \sqrt{5}$

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(B) The given equation of the ellipse is $9x^2+4y^2-18x-8y-23=0$. Completing the square,we get: $9(x^2-2x+1) + 4(y^2-2y+1) = 23+9+4$ $9(x-1)^2 + 4(y-1)^2 = 36$ Dividing by $36$,we get: $\frac{(x-1)^2}{4} + \frac{(y-1)^2}{9} = 1$. Here,$a^2=4$ and $b^2=9$. Since $b^2 > a^2$,the major axis is vertical. The eccentricity $e = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}$. The foci are given by $(h, k \pm be)$,where $(h, k) = (1, 1)$. Foci $= (1, 1 \pm 3 	imes \frac{\sqrt{5}}{3}) = (1, 1 \pm \sqrt{5})$. The equations of the latus rectum are $y = k \pm be$,which simplifies to $y = 1 \pm \sqrt{5}$.

The equations of the latus rectum of the ellipse $9x^2+4y^2-18x-8y-23=0$ are

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