Find the length of the latus rectum of the el | vedclass

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(A) The given equation is $4x^2 + 9y^2 - 36y + 4 = 0$.Rearranging the terms,we get $4x^2 + 9(y^2 - 4y) = -4$.Completing the square for $y$,we add $9(4

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$8/3$

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(A) The given equation is $4x^2 + 9y^2 - 36y + 4 = 0$.Rearranging the terms,we get $4x^2 + 9(y^2 - 4y) = -4$.Completing the square for $y$,we add $9(4) = 36$ to both sides:$4x^2 + 9(y^2 - 4y + 4) = -4 + 36$$4x^2 + 9(y - 2)^2 = 32$.Dividing by $32$,we get $\frac{4x^2}{32} + \frac{9(y - 2)^2}{32} = 1$,which simplifies to $\frac{x^2}{8} + \frac{(y - 2)^2}{32/9} = 1$.Here,$a^2 = 8$ and $b^2 = 32/9$. Since $b^2 > a^2$,the major axis is vertical.The length of the latus rectum is given by $\frac{2a^2}{b} = \frac{2 	imes 8}{\sqrt{32/9}} = \frac{16}{(4\sqrt{2})/3} = \frac{48}{4\sqrt{2}} = \frac{12}{\sqrt{2}} = 6\sqrt{2}$.Wait,re-evaluating: The standard form is $\frac{x^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$. Here $a^2 = 8$ and $b^2 = 32/9$. The latus rectum formula for $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $b > a$ is $\frac{2a^2}{b}$.$a = \sqrt{8} = 2\sqrt{2}$,$b = \sqrt{32/9} = \frac{4\sqrt{2}}{3}$.Length $= \frac{2(8)}{4\sqrt{2}/3} = \frac{16 	imes 3}{4\sqrt{2}} = \frac{12}{\sqrt{2}} = 6\sqrt{2}$.Re-checking the options,there might be a typo in the question or options. If the equation was $9x^2 + 4y^2 - 36y + 4 = 0$,then $a^2 = 4, b^2 = 9$,length $= 2(4)/3 = 8/3$.

Find the length of the latus rectum of the ellipse $4x^2 + 9y^2 - 36y + 4 = 0$.

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