Find the vertex, focus, and directrix of the parabola. $y ^ { 2 } = - 3 x$
A)Vertex: (0,0); Focus: $\left( - \frac { 3 } { 4 } , 0 \right)$ ; Directrix: $x = - \frac { 3 } { 4 }$
B)Vertex: (0,0); Focus: $\left( - \frac { 1 } { 3 } , 0 \right)$ ; Directrix: $x = \frac { 1 } { 3 }$
C)Vertex: (0,0); Focus: $\left( \frac { 1 } { 3 } , 0 \right)$ ; Directrix: $x = \frac { 1 } { 3 }$
D)Vertex: (0,0); Focus: $\left( \frac { 3 } { 4 } , 0 \right)$ ; Directrix: $x = - \frac { 3 } { 4 }$
E)Vertex: (0,0); Focus: $\left( - \frac { 3 } { 4 } , 0 \right)$ ; Directrix: $x = \frac { 3 } { 4 }$

Question

Find the vertex, focus, and directrix of the parabola.  $y ^ { 2 } = - 3 x$ 
A)Vertex: (0,0); Focus: $\left( - \frac { 3 } { 4 } , 0 \right)$ ; Directrix: $x = - \frac { 3 } { 4 }$
B)Vertex: (0,0); Focus: $\left( - \frac { 1 } { 3 } , 0 \right)$ ; Directrix: $x = \frac { 1 } { 3 }$
C)Vertex: (0,0); Focus: $\left( \frac { 1 } { 3 } , 0 \right)$ ; Directrix: $x = \frac { 1 } { 3 }$
D)Vertex: (0,0); Focus: $\left( \frac { 3 } { 4 } , 0 \right)$ ; Directrix: $x = - \frac { 3 } { 4 }$
E)Vertex: (0,0); Focus: $\left( - \frac { 3 } { 4 } , 0 \right)$ ; Directrix: $x = \frac { 3 } { 4 }$

Quizplus · Accepted Answer

The equation $y^2 = -3x$ represents a parabola that opens to the left. The general form of a parabola opening to the left or right is $y^2 = 4px$, where $p$ is the distance from the vertex to the focus and also from the vertex to the directrix. Here, $4p = -3$, so $p = -\frac{3}{4}$. The vertex is at the origin (0,0), the focus is $p$ units to the left of the vertex at $\left(-\frac{3}{4}, 0\right)$, and the directrix is a vertical line $p$ units to the right of the vertex, which is $x = \frac{3}{4}$.

Find the Vertex, Focus, and Directrix of the Parabola y2=−3xy ^ { 2 } = - 3 xy2=−3x

Find the Vertex, Focus, and Directrix of the Parabola $y ^ { 2 } = - 3 x$