Question

The length of latus rectum of the parabola $4y^2+3x+3y+1=0$ is

• A. $-\frac{3}{4}$
• B. $7$
• C. $12$
• D. $\frac{3}{4}$

Answer 1

Answer: (D)

$\frac{3}{4}$

$4y^2+3x+3y+1=0$

$\Rightarrow 4y^2+3y=-3x-1$ $,\text{take 3x+1 to RHS}$

$\Rightarrow y^2+\frac{3}{4}y=-\frac{3}{4}x-\frac{1}{4}$ $,\text{divide each sides by 4}$

$\Rightarrow y^2+2\cdot \frac{3}{8}y+{\left(\frac{3}{8}\right)}^2=-\frac{3}{4}x-\frac{1}{4}+{\left(\frac{3}{8}\right)}^2$ $,\text{add each sides}{\left(\frac{3}{8}\right)}^2$

$\Rightarrow {(y+\frac{3}{8})}^2=-\frac{3}{4}x-\frac{1}{4}+\frac{9}{64}=-\frac{3}{4}x-\frac{15}{64}$

$\Rightarrow {(y+\frac{3}{8})}^2=-\frac{3}{4}(x+\frac{5}{16})$

Now comparing above equation with standard parabola $Y^2=-4aX$

Length of latus rectum $=4a=\frac{3}{4}$