The length of the latus rectum of the parabola $3 x^{2} + 4 y + 5 + 6 x = 0$ is

\frac{1}{3}

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\frac{2}{3}

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\frac{5}{3}

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\frac{4}{3}

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Solution

The correct option is D $\frac{4}{3}$
The equation of the parabola is,
$3 x^{2} + 4 y + 5 + 6 x = 0 \Rightarrow 3 x^{2} + 6 x + 5 = - 4 y \Rightarrow 3 (x^{2} + 2 x + 1) + 2 = - 4 y \Rightarrow 3 (x + 1)^{2} = - 4 (y + \frac{1}{2}) \Rightarrow (x + 1)^{2} = - \frac{4}{3} (y + \frac{1}{2})$

Comparing with the standard form $x^{2} = - 4 a y$
The lenght of latus rectum $= 4 a = \frac{4}{3}$

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