Question

a parabola with a vertex at (0,0) has a directrix that crosses the negative part of the y - axis. which could be the equation of the parabola? $x^2 = - 4y$; $x^2 = 4y$; $y^2 = 4x$; $y^2 = - 4x$

Explanation:

Step1: Recall parabola standard forms

For a parabola with vertex at \((0,0)\):
- If it opens up/down, the equation is \(x^2 = 4py\) (opens up if \(p>0\), down if \(p<0\)). The directrix is \(y=-p\).
- If it opens left/right, the equation is \(y^2 = 4px\) (opens right if \(p>0\), left if \(p<0\)). The directrix is \(x=-p\).

Step2: Analyze direction of opening

The directrix crosses the negative \(y\)-axis, so the parabola opens along the \(y\)-axis (eliminates \(y^2 = 4x\) and \(y^2=-4x\) as they open along \(x\)-axis).

Step3: Determine \(p\) sign

For \(x^2 = 4py\), directrix \(y=-p\). Negative \(y\)-axis directrix means \(y=-p < 0\) (since directrix is on negative \(y\)-axis), so \(-p < 0\Rightarrow p>0\)? Wait, no: Wait, if directrix is \(y = -p\), and it's on negative \(y\)-axis, then \(-p < 0\Rightarrow p>0\)? Wait, no, let's check the options:
- For \(x^2 = 4y\): \(4p = 4\Rightarrow p = 1\), directrix \(y=-1\) (negative \(y\)-axis? Wait, \(y=-1\) is negative \(y\)-axis? Wait, no, \(y=-1\) is below \(y=0\), so negative \(y\)-axis? Wait, but wait, the problem says "crosses the negative part of the \(y\)-axis". Wait, maybe I mixed up. Wait, if the parabola opens up, directrix is below the vertex (at \(y=-p\)). If \(p>0\), directrix \(y=-p\) is negative. Wait, but let's check the options:
- \(x^2 = 4y\): \(p = 1\), directrix \(y=-1\) (negative \(y\)-axis? Wait, \(y=-1\) is negative. But wait, the other option \(x^2=-4y\): \(4p=-4\Rightarrow p=-1\), directrix \(y = -p = 1\) (positive \(y\)-axis). Wait, I think I made a mistake earlier. Let's re-express:

Standard form \(x^2 = 4py\):
- If \(p>0\), opens up, directrix \(y=-p\) (negative \(y\)-axis when \(p>0\)? Wait, no: \(y=-p\) when \(p>0\) is negative (e.g., \(p=1\), \(y=-1\)). Wait, but the problem says "directrix that crosses the negative part of the \(y\)-axis". So directrix is a horizontal line (since it's a vertical parabola) crossing negative \(y\)-axis, so directrix \(y = k\) where \(k < 0\). For \(x^2 = 4py\), directrix \(y=-p\). So \(y=-p < 0\Rightarrow p>0\)? Wait, no: if \(p>0\), then \(y=-p\) is negative. Wait, but let's check the options:

- \(x^2 = 4y\): \(p = 1\), directrix \(y=-1\) (negative \(y\)-axis? Wait, \(y=-1\) is on the negative \(y\)-axis? Wait, the \(y\)-axis is vertical, so a horizontal line (directrix for vertical parabola) crossing the negative \(y\)-axis means the directrix is \(y = -k\) where \(k>0\), so the parabola opens up (since directrix is below vertex). Wait, but the other option \(x^2=-4y\): \(p=-1\), directrix \(y = 1\) (positive \(y\)-axis). So between \(x^2=4y\) and \(x^2=-4y\), \(x^2=4y\) has directrix \(y=-1\) (negative \(y\)-axis), \(x^2=-4y\) has directrix \(y=1\) (positive \(y\)-axis). The problem says directrix crosses negative part of \(y\)-axis, so \(x^2=4y\) is the one? Wait, no, wait: Wait, the vertex is at \((0,0)\). If the directrix is on the negative \(y\)-axis, the parabola opens towards the positive \(y\)-axis (since parabola is equidistant from focus and directrix; focus is on the opposite side of directrix from the parabola's opening). So if directrix is \(y=-p\) (negative \(y\)-axis), then focus is at \((0,p)\), so parabola opens up (towards positive \(y\)-axis), so equation is \(x^2 = 4py\) with \(p>0\). So \(x^2=4y\) (since \(4p=4\Rightarrow p=1>0\)) has directrix \(y=-1\) (negative \(y\)-axis), which matches. Wait, but earlier I thought maybe I messed up, but let's confirm:

- \(x^2=4y\): opens up, vertex \((0,0)\), focus \((0,1)\), directrix \(y=-1\) (crosses negative \(y\)-axis).
- \(x^2=-4y\): opens down, directrix \(y=1\) (positive \(y\)-axis).
- \(y^2=4x\): opens right, directrix \(x=-1\) (vertical line, crosses negative \(x\)-axis, not \(y\)-axis).
- \(y^2=-4x\): opens left, directrix \(x=1\) (positive \(x\)-axis).

So the correct equation is \(x^2=4y\)? Wait, wait, no: Wait, the problem says "directrix that crosses the negative part of the \(y\)-axis". A horizontal line (directrix for vertical parabola) crossing the negative \(y\)-axis: the directrix is \(y = -k\) where \(k>0\), so the parabola opens up (towards positive \(y\)-axis), so equation \(x^2=4py\) with \(p>0\), so \(x^2=4y\) (p=1) has directrix \(y=-1\) (negative \(y\)-axis), which is correct. Wait, but let me check again. Wait, maybe I made a mistake in the sign. Let's recall:

For parabola \(x^2 = 4py\):
- If \(p > 0\), opens upward, focus \((0, p)\), directrix \(y = -p\) (below the vertex).
- If \(p < 0\), opens downward, focus \((0, p)\), directrix \(y = -p\) (above the vertex).

So if directrix is on the negative \(y\)-axis, \(y = -p < 0\) (since it's on negative \(y\)-axis), so \(-p < 0\Rightarrow p > 0\). Therefore, \(x^2 = 4py\) with \(p > 0\), so \(x^2 = 4y\) (p=1) is correct. Wait, but the option \(x^2=4y\) is one of the choices. Wait, but let's check the other options:

- \(x^2=-4y\): \(p=-1\), directrix \(y=1\) (positive \(y\)-axis) – no.
- \(y^2=4x\): opens right, directrix \(x=-1\) (negative \(x\)-axis) – no.
- \(y^2=-4x\): opens left, directrix \(x=1\) (positive \(x\)-axis) – no.

So the correct equation is \(x^2=4y\)? Wait, but wait, the problem says "directrix that crosses the negative part of the \(y\)-axis". A horizontal line (directrix) crossing the negative \(y\)-axis: the line \(y = -1\) (directrix of \(x^2=4y\)) crosses the \(y\)-axis at \((0, -1)\), which is on the negative \(y\)-axis. Yes, that's correct. So the equation is \(x^2=4y\). Wait, but wait, I think I made a mistake earlier. Wait, no: Wait, if the directrix is on the negative \(y\)-axis, the parabola opens towards the positive \(y\)-axis, so the equation is \(x^2 = 4py\) with \(p>0\), so \(x^2=4y\) (p=1) is correct.

Answer:

\(\boldsymbol{x^2 = 4y}\) (Option: \(x^2 = 4y\))