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

Given parabola
\[
\begin{array}{l}
4 y^{2}+3 x+3 y+1=0 \\
\Rightarrow y^{2}+\frac{3}{4} y+\frac{1}{4}=\frac{-3}{4} x \\
\Rightarrow\left(y+\frac{3}{8}\right)^{2}-\left(\frac{3}{8}\right)^{2}+\frac{1}{4}=\frac{-3}{4} x \\
\Rightarrow\left(y+\frac{3}{8}\right)^{2}+\frac{7}{64}=\frac{-3}{4} x \\
\Rightarrow\left(y+\frac{3}{8}\right)^{2}=-\frac{3}{4} x-\frac{7}{64} \\
\Rightarrow\left(y+\frac{3}{8}\right)^{2}=\frac{-3}{4}\left(x+\frac{7}{48}\right) \\
\Rightarrow\left(y+\frac{3}{8}\right)^{2}=\frac{-3}{4}\left(x+\frac{7}{48}\right) \rightarrow(1)
\end{array}
\]
Comparing Eq. (1) with standard equation of parabola,
\( y^{2}=4 a X \)
Therefore, length of latus rectum is \( \frac{3}{4} \)