The length of the latus rectum of the parabol | vedclass

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Question

(D) Given the equation of the parabola: $4y^{2} + 3x + 3y + 1 = 0$. Rearranging the terms to isolate $y$ terms: $4y^{2} + 3y = -3x - 1$. Dividing by $

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$\frac{3}{4}$

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(D) Given the equation of the parabola: $4y^{2} + 3x + 3y + 1 = 0$. Rearranging the terms to isolate $y$ terms: $4y^{2} + 3y = -3x - 1$. Dividing by $4$: $y^{2} + \frac{3}{4}y = -\frac{3}{4}x - \frac{1}{4}$. Completing the square on the left side: $(y + \frac{3}{8})^{2} - \frac{9}{64} = -\frac{3}{4}x - \frac{1}{4}$. $(y + \frac{3}{8})^{2} = -\frac{3}{4}x - \frac{1}{4} + \frac{9}{64}$. $(y + \frac{3}{8})^{2} = -\frac{3}{4}x - \frac{16}{64} + \frac{9}{64}$. $(y + \frac{3}{8})^{2} = -\frac{3}{4}x - \frac{7}{64}$. $(y + \frac{3}{8})^{2} = -\frac{3}{4}(x + \frac{7}{48})$. Comparing this with the standard form $(y - k)^{2} = -4a(x - h)$,we get $4a = \frac{3}{4}$. Thus,the length of the latus rectum is $4a = \frac{3}{4}$.

The length of the latus rectum of the parabola $4y^{2} + 3x + 3y + 1 = 0$ is

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