Equations of the latus rectum of the ellipse | vedclass

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(C) 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) - 9 + 4(y^2 - 2y + 1) - 4 -

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

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(C) The given equation of the ellipse is $9x^2 + 4y^2 - 18x - 8y - 23 = 0$. Completing the square,we get: $9(x^2 - 2x + 1) - 9 + 4(y^2 - 2y + 1) - 4 - 23 = 0$ $9(x - 1)^2 + 4(y - 1)^2 = 36$ Dividing by $36$,we obtain the standard form: $\frac{(x - 1)^2}{4} + \frac{(y - 1)^2}{9} = 1$ Here,$a^2 = 4$ and $b^2 = 9$,so $a^2 The eccentricity $e$ is given by $e = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}$. The equations of the latus rectum for a vertical ellipse are $y - k = \pm be$,where $(h, k) = (1, 1)$. $y - 1 = \pm 3 	imes \frac{\sqrt{5}}{3}$ $y - 1 = \pm \sqrt{5}$ $y = 1 \pm \sqrt{5}$

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

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