The length of the latus rectum of 3x^{2} - 4y | vedclass

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(B) Given equation of the conic is $3x^{2} - 4y + 6x - 3 = 0$. Rearranging the terms,we get $3x^{2} + 6x = 4y + 3$. Taking $3$ as a common factor,$3(x

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$\frac{4}{3}$

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(B) Given equation of the conic is $3x^{2} - 4y + 6x - 3 = 0$. Rearranging the terms,we get $3x^{2} + 6x = 4y + 3$. Taking $3$ as a common factor,$3(x^{2} + 2x) = 4y + 3$. Completing the square inside the bracket,$3(x^{2} + 2x + 1 - 1) = 4y + 3$. $3((x + 1)^{2} - 1) = 4y + 3$. $3(x + 1)^{2} - 3 = 4y + 3$. $3(x + 1)^{2} = 4y + 6$. $(x + 1)^{2} = \frac{4}{3}(y + \frac{6}{4}) = \frac{4}{3}(y + \frac{3}{2})$. Comparing this with the standard form of a parabola $X^{2} = 4aY$,where $X = x + 1$ and $Y = y + \frac{3}{2}$,we have $4a = \frac{4}{3}$. The length of the latus rectum is $4a = \frac{4}{3}$.

The length of the latus rectum of $3x^{2} - 4y + 6x - 3 = 0$ is

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