Question

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

• A. $2y+3=0$
• B. $3y=2$
• C. $2y=3$
• D. $3y+2=0$

Answer 1

The correct option is C $2y=3$

$x^2+4x+2y=0\to x^2+4x+4+2y-4=0\to (x+2{)}^2=2(2-y)$

Now, focus of parabola $(x-h{)}^2=4p(y-k)$ is $F(h,k+p)$

Here $h=-2,p=-\frac{1}{2},k=2\to F(-2,\frac{3}{2})$

Equation of axis is $x-h=0\to x+2=0$

Now latus rectum is perpendicular to axis and passes through $F$.

$\therefore$Slope of latus rectum$=0$

Equation will be $y-\frac{3}{2}=0(x-(-2\left)\right)\to y=\frac{3}{2}\to 2y=3$

Hence, $\left(C\right)$.