Question

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

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

Answer 1

The correct option is D $\frac{3}{4}$

$4y^2+3x+3y+1=0$
$4(y^2+\frac{3}{4}y)=-3x-1$
$y^2+\frac{3}{4}y=-\frac{3}{4}x-\frac{1}{4}$
$y^2+\frac{3}{4}y+\frac{9}{64}=-\frac{3}{4}x-\frac{1}{4}+\frac{9}{64}$
$(y+\frac{3}{8}{)}^2=-\frac{3}{4}x-\frac{7}{64}$
${(y+\frac{3}{8})}^2=-\frac{3}{4}(x+\frac{7}{48})$
Comparing to $Y^2=4aX$
we get $4a=-\frac{3}{4}$
Length of latus rectum $=\frac{3}{4}$