Q81 / 180 · KCET 2016 · Mathematics
The length of latus rectum of the parabola \( 4 y^{2}+3 x+3 y+1=0 \) is
- A\( \frac{4}{3} \)
- B\( 07 \)
- C\( 12 \)
- Correct answerD\( \frac{3}{4} \)
Solution
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 \( \frac{3}{4} \)
\[ \begin{array}{l} 4 y^{2}+3 x+3 y+1=0 \\ \Rightarrow y^{2}+\frac{3}{4} y+\frac{1}{4}=\frac{-3}{4} x \\ \Rightarrow\left(y+\frac{3}{8}\right)^{2}-\left(\frac{3}{8}\right)^{2}+\frac{1}{4}=\frac{-3}{4} x \\ \Rightarrow\left(y+\frac{3}{8}\right)^{2}+\frac{7}{64}=\frac{-3}{4} x \\ \Rightarrow\left(y+\frac{3}{8}\right)^{2}=-\frac{3}{4} x-\frac{7}{64} \\ \Rightarrow\left(y+\frac{3}{8}\right)^{2}=\frac{-3}{4}\left(x+\frac{7}{48}\right) \\ \Rightarrow\left(y+\frac{3}{8}\right)^{2}=\frac{-3}{4}\left(x+\frac{7}{48}\right) \rightarrow(1) \end{array} \]
Comparing Eq. (1) with standard equation of parabola,
\( y^{2}=4 a X \)
Therefore, length of latus rectum is \( \frac{3}{4} \)
Hence, the correct answer is option: 'D'.