Find the vertex, and directrix of the parabola. $y ^ { 2 } + 2 y - 12 x + 37 = 0$
A) $V ( 6 , - 1 ) , x = 0$
B) $V ( 3 , - 1 )$ , $x = 0$
C) $V ( 6 , - 1 ) , y = 0$
D) $V ( 4 , - 1 ) , y = - 1$
E) $V ( 6,0 ) , x = - 1$

Question

Quizplus · Accepted Answer

We can rewrite the given equation as $y^2+2y+1=12x-26+1$ or $(y+1)^2=12(x-3)$. Therefore, the vertex is $(3,-1)$ and the axis of symmetry is $x=3$. The distance between the vertex and the directrix is $\dfrac{1}{4p}=3$, where $p$ is the distance between the vertex and the focus. Solving for $p$, we get $p=\dfrac{1}{12}$, so the directrix is $\boxed{	extbf{(B) }x=0}$.

Find the Vertex, and Directrix of the Parabola y2+2y−12x+37=0y ^ { 2 } + 2 y - 12 x + 37 = 0y2+2y−12x+37=0

Find the Vertex, and Directrix of the Parabola $y ^ { 2 } + 2 y - 12 x + 37 = 0$