The length of the latus rectum of the parabol | vedclass

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(D) The given equation is $9x^2 + 24xy + 16y^2 - 4x + 3y = 0$.This can be written as $(3x + 4y)^2 - 4x + 3y = 0$.Let $L = 3x + 4y$ and $M = 4x - 3y$.

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$1$

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(D) The given equation is $9x^2 + 24xy + 16y^2 - 4x + 3y = 0$.This can be written as $(3x + 4y)^2 - 4x + 3y = 0$.Let $L = 3x + 4y$ and $M = 4x - 3y$. Note that $L$ and $M$ are perpendicular lines since their slopes are $-3/4$ and $4/3$ respectively.We normalize the lines: $\left(\frac{3x + 4y}{5}\right)^2 = \frac{4x - 3y}{5}$.This is of the form $Y^2 = 4aX$,where $Y = \frac{3x + 4y}{5}$ and $X = \frac{4x - 3y}{5}$.Here,$4a = 1$,so $a = \frac{1}{4}$.The length of the latus rectum is $4a = 1$.

The length of the latus rectum of the parabola $9x^2 + 16y^2 + 24xy - 4x + 3y = 0$ is

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