Parabola Equation: Focus At (-2, 0)

Hey guys, let's dive into the awesome world of parabolas! Today, we're tackling a super common problem: finding the equation of a parabola when you know where its focus is. Specifically, we're going to figure out the equation of a parabola with its focus at (-2, 0). This is a fundamental concept in understanding conic sections, and once you get the hang of it, you'll see parabolas everywhere – from satellite dishes to the trajectory of a thrown ball. So, grab your notebooks, and let's break it down!

Understanding the Basics of a Parabola

Before we jump into our specific problem, let's quickly recap what makes a parabola a parabola. A parabola is a set of all points in a plane that are the same distance from a fixed point (called the focus) and a fixed line (called the directrix). This elegant definition is the key to deriving its equation. The vertex of the parabola is the midpoint between the focus and the directrix, and it lies on the axis of symmetry, which passes through the focus and is perpendicular to the directrix.

For our problem, the focus is given as F(-2, 0). This tells us a few important things. Since the y-coordinate of the focus is 0, the axis of symmetry must be the x-axis (y=0). This means our parabola will open either to the left or to the right. If the focus were (0, k), the axis of symmetry would be the y-axis (x=0), and the parabola would open up or down.

Deriving the Standard Equation

There are a couple of standard forms for parabola equations. When the axis of symmetry is horizontal (like in our case, the x-axis), the standard equation is of the form (y - k)^2 = 4p(x - h). Here, (h, k) represents the coordinates of the vertex, and p is the directed distance from the vertex to the focus (and also from the vertex to the directrix, but in the opposite direction).

In our problem, the focus is at (-2, 0). Since the axis of symmetry is the x-axis, the vertex must lie on the x-axis as well. Let's assume the vertex is at (h, 0). The distance from the vertex (h, 0) to the focus (-2, 0) is $|-2 - h|$. This distance is equal to $|p|$.

Now, how do we find 'p' and the vertex (h, k)? The definition of a parabola is crucial here. Any point (x, y) on the parabola is equidistant from the focus F(-2, 0) and the directrix. Let's denote the directrix as the line $x = d$. The distance from (x, y) to F(-2, 0) is $\sqrt{(x - (-2))^2 + (y - 0)^2} = \sqrt{(x + 2)^2 + y^2}$. The distance from (x, y) to the line $x = d$ is $|x - d|$.

Setting these equal, we get: $\sqrt{(x + 2)^2 + y^2} = |x - d|$. Squaring both sides: $(x + 2)^2 + y^2 = (x - d)^2$. Expanding: $x^2 + 4x + 4 + y^2 = x^2 - 2dx + d^2$. Simplifying by canceling $x^2$: $4x + 4 + y^2 = -2dx + d^2$. Rearranging to group x terms: $(4 + 2d)x + y^2 = d^2 - 4$.

This equation still has 'd' in it. We need to relate 'd' to the vertex (h, k). We know the vertex is (h, k), and for a horizontal parabola, k = 0. So, the vertex is (h, 0).

The distance from the vertex (h, 0) to the focus (-2, 0) is $|-2 - h|$. This is our 'p'. The directrix is $x = d$. The vertex (h, 0) is exactly halfway between the focus (-2, 0) and the directrix $x = d$. Therefore, $h = \frac{-2 + d}{2}$. This means $2h = -2 + d$, or $d = 2h + 2$.

Also, the directed distance $p$ from the vertex to the focus is $-2 - h$. So, $p = -2 - h$. This implies $h = -2 - p$. Substituting $d = 2h + 2$ into the simplified equation $(4 + 2d)x + y^2 = d^2 - 4$: $(4 + 2(2h + 2))x + y^2 = (2h + 2)^2 - 4$ $(4 + 4h + 4)x + y^2 = (4h^2 + 8h + 4) - 4$ $(4h + 8)x + y^2 = 4h^2 + 8h$ $4(h + 2)x + y^2 = 4h(h + 2)$

Now, let's use the relationship $p = -2 - h$, which means $h + 2 = -p$. Substituting this back: $4(-p)x + y^2 = 4h(-p)$ $-4px + y^2 = -4hp$

This doesn't look like the standard form yet. Let's go back to the equation $(x + 2)^2 + y^2 = (x - d)^2$ and use the vertex information more directly.

We know the vertex is (h, k). For a parabola opening horizontally, the equation is $(y - k)^2 = 4p(x - h)$. The focus is at (h + p, k). We are given the focus is (-2, 0). So, we can equate these: h + p = -2 k = 0

This gives us two equations. Since k = 0, our vertex is (h, 0). The equation becomes $(y - 0)^2 = 4p(x - h)$, which simplifies to $y^2 = 4p(x - h)$.

We have one equation, $h + p = -2$, but two unknowns, $h$ and $p$. This means there isn't a unique parabola with just the focus given. We need more information, such as the vertex or the directrix. However, the standard form implies that 'p' is the distance from the vertex to the focus. If we assume the vertex is at the origin (0, 0), then $h=0$ and $k=0$. In this special case, the focus would be $(0+p, 0) = (p, 0)$. If the focus is (-2, 0), then $p = -2$. The equation would be $y^2 = 4(-2)(x - 0)$, which is $y^2 = -8x$. This parabola has its vertex at the origin and focus at (-2, 0). Its directrix would be $x = -p = 2$. Let's check: distance from any point (x,y) on $y^2 = -8x$ to focus (-2,0) is $\sqrt{(x+2)^2 + y^2} = \sqrt{(x+2)^2 - 8x} = \sqrt{x^2 + 4x + 4 - 8x} = \sqrt{x^2 - 4x + 4} = \sqrt{(x-2)^2} = |x-2|$. The distance to the directrix $x=2$ is $|x-2|$. They match!

But what if the vertex isn't at the origin? The problem statement doesn't specify the vertex. Let's re-examine the structure of the standard equation $(y - k)^2 = 4p(x - h)$. The focus is always at $(h+p, k)$. We are given that the focus is $(-2, 0)$. So, we have: $h + p = -2$ $k = 0$

The equation of the parabola is therefore $(y - 0)^2 = 4p(x - h)$, which is $y^2 = 4p(x - h)$.

We can express $h$ in terms of $p$: $h = -2 - p$. Substituting this into the equation: $y^2 = 4p(x - (-2 - p))$ $y^2 = 4p(x + 2 + p)$

This is the general form of a parabola with focus at (-2, 0). Notice that there are infinitely many such parabolas, each defined by a different value of 'p' (which determines the vertex). For example:

• If $p = -1$, then $h = -2 - (-1) = -1$. The vertex is (-1, 0). The equation is $y^2 = 4(-1)(x - (-1)) = -4(x + 1)$.
• If $p = -2$, then $h = -2 - (-2) = 0$. The vertex is (0, 0). The equation is $y^2 = 4(-2)(x - 0) = -8x$. (This is the case we found earlier).
• If $p = -3$, then $h = -2 - (-3) = 1$. The vertex is (1, 0). The equation is $y^2 = 4(-3)(x - 1) = -12(x - 1)$.

As you can see, the value of 'p' determines how