Question

the parabola has a focus at (-3, 0) and directrix x = 3. what is the correct equation for the parabola? $\circ\\ x^2 = -12y$ $\circ\\ x^2 = 3y$ $\circ\\ y^2 = 3x$ $\circ\\ y^2 = -12x$

Explanation:

Step1: Determine the orientation and vertex

The focus is at \((-3, 0)\) and the directrix is \(x = 3\). Since the directrix is a vertical line and the focus is on the x - axis, the parabola opens horizontally. The vertex \((h,k)\) is the mid - point between the focus \((-3,0)\) and the directrix \(x = 3\). The x - coordinate of the vertex \(h=\frac{-3 + 3}{2}=0\), and the y - coordinate \(k = 0\). So the vertex is \((0,0)\).

Step2: Find the value of \(p\)

For a horizontally opening parabola, the standard form is \((y - k)^2=4p(x - h)\). The distance between the focus and the vertex is \(|p|\). The focus is at \((-3,0)\) and the vertex is at \((0,0)\), so \(p=-3\) (negative because the parabola opens to the left).

Step3: Write the equation of the parabola

Substitute \(h = 0\), \(k = 0\) and \(p=-3\) into the standard form \((y - k)^2=4p(x - h)\). We get \((y-0)^2=4\times(-3)(x - 0)\), which simplifies to \(y^{2}=-12x\).

Answer:

\(y^{2}=-12x\) (the option: \(y^{2}=-12x\))