The equations of the latus rectum of the elli | vedclass

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(B) Given equation: $9x^2 + 25y^2 - 36x + 50y - 164 = 0$ Rearranging terms: $9(x^2 - 4x) + 25(y^2 + 2y) = 164$ Completing the square: $9(x^2 - 4x + 4)

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$x-6=0, x+2=0$

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(B) Given equation: $9x^2 + 25y^2 - 36x + 50y - 164 = 0$ Rearranging terms: $9(x^2 - 4x) + 25(y^2 + 2y) = 164$ Completing the square: $9(x^2 - 4x + 4) + 25(y^2 + 2y + 1) = 164 + 36 + 25$ $9(x-2)^2 + 25(y+1)^2 = 225$ Dividing by $225$: $\frac{(x-2)^2}{25} + \frac{(y+1)^2}{9} = 1$ Here,$a^2 = 25$ and $b^2 = 9$,so $a = 5$ and $b = 3$. Eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$. The equations of the latus rectum are $x - h = \pm ae$. $x - 2 = \pm 5 	imes \frac{4}{5} = \pm 4$. $x = 2 + 4 = 6$ and $x = 2 - 4 = -2$. Thus,the equations are $x - 6 = 0$ and $x + 2 = 0$.

The equations of the latus rectum of the ellipse $9x^2 + 25y^2 - 36x + 50y - 164 = 0$ are

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