Question

which equation represents the parabola shown on the graph?

• $y^2 = -2x$
• $y^2 = -8x$
• $x^2 = -2y$
• $x^2 = -8y$

Explanation:

Step1: Determine the orientation and vertex

The parabola opens to the left (since it's symmetric about the x - axis and opens in the negative x - direction) and has its vertex at \((- 2,0)\). The standard form of a parabola that opens left or right is \(y^{2}=4p(x - h)\), where \((h,k)\) is the vertex. Here, \(h=-2\), \(k = 0\), so the equation is \(y^{2}=4p(x + 2)\). Also, the directrix is \(x = 2\). For a parabola of the form \(y^{2}=4p(x - h)\), the distance between the vertex \((h,k)\) and the directrix \(x=h - p\) (if it opens left) or \(x=h + p\) (if it opens right). Since it opens left, the directrix is \(x=h - p\). We know \(h=-2\) and the directrix \(x = 2\), so \(2=-2 - p\), solving for \(p\) gives \(p=-4\).

Step2: Substitute p into the standard form

Substitute \(p=-4\) and \((h,k)=(-2,0)\) into the standard form \(y^{2}=4p(x - h)\). We get \(y^{2}=4\times(-4)(x + 2)=-16(x + 2)\)? Wait, no, maybe a better way. Let's check the options. The options are in the form \(y^{2}=-2x\), \(y^{2}=-8x\), \(x^{2}=-2y\), \(x^{2}=-8y\). The parabola is symmetric about the x - axis (since for every point \((x,y)\) on the parabola, \((x, - y)\) is also on it), so the equation should be of the form \(y^{2}=4p x\) (vertex at the origin? Wait, maybe I made a mistake in the vertex. Wait, the vertex is at \((-2,0)\), but let's check the options. The options have vertex at the origin? Wait, no, maybe the graph is shifted? Wait, no, looking at the graph, when \(y = 0\), \(x=-2\), but the options are \(y^{2}=-2x\), \(y^{2}=-8x\) (vertex at \((0,0)\)), \(x^{2}=-2y\), \(x^{2}=-8y\) (vertex at \((0,0)\)). Wait, maybe the vertex is at \((0,0)\) and I misread the graph. Wait, the point \((-2,0)\) is on the parabola? Wait, no, the vertex is at \((-2,0)\), but the options are centered at the origin. Wait, maybe the graph is actually with vertex at \((0,0)\) and I misread. Let's check the symmetry. The parabola is symmetric about the x - axis, so it's a horizontal parabola, so the equation is of the form \(y^{2}=4p x\). The directrix of \(y^{2}=4p x\) is \(x=-p\). If the directrix is \(x = 2\), then \(-p=2\), so \(p=-2\)? No, wait, for \(y^{2}=4p x\), if it opens left, \(p<0\), and the directrix is \(x=-p\). Wait, if the directrix is \(x = 2\), then \(-p = 2\), so \(p=-2\), then \(4p=-8\), so the equation is \(y^{2}=-8x\). Let's check a point. Take \(x=-2\), then \(y^{2}=-8\times(-2)=16\), so \(y=\pm4\). Looking at the graph, when \(x=-2\), \(y\) is around \(\pm4\) (since at \(x=-2\), the graph is at \(y = 4\) and \(y=-4\) approximately). Let's check \(y^{2}=-8x\): when \(x=-2\), \(y^{2}=16\), \(y=\pm4\), which matches the graph. For \(y^{2}=-2x\), when \(x=-2\), \(y^{2}=4\), \(y=\pm2\), which does not match the graph (the graph at \(x=-2\) has \(y\) values of \(\pm4\)). The other two options are vertical parabolas (symmetric about y - axis), so they can be eliminated.

Answer:

\(y^{2}=-8x\) (the option is \(y^{2}=-8x\))