Finding The Equation Of A Parabola Focus (-3,0) And Directrix X=3

In the realm of conic sections, the parabola stands out as a fascinating curve with numerous applications in physics, engineering, and mathematics. Defining a parabola hinges on two key elements: the focus, a fixed point, and the directrix, a fixed line. This exploration delves into the process of determining the equation of a parabola given its focus and directrix. Specifically, we will tackle the problem of finding the equation of a parabola with a focus at (-3, 0) and a directrix of x = 3. Understanding the fundamental definition of a parabola and how it relates to the focus and directrix is paramount to solving this problem. The definition of a parabola states that it is the set of all points that are equidistant from the focus and the directrix. This geometric property forms the basis for deriving the algebraic equation of the parabola. By applying the distance formula and algebraic manipulation, we can transform this geometric definition into a concrete equation that describes the parabola's shape and position in the coordinate plane. This article will guide you through the steps necessary to derive the equation, providing a clear understanding of the relationship between the focus, directrix, and the resulting parabolic equation. We will explore the standard forms of parabolic equations and how the position of the focus and directrix influence the equation's coefficients. Furthermore, this process reinforces the crucial link between geometric properties and algebraic representations, a cornerstone of analytic geometry. The ability to translate geometric information into algebraic equations is a fundamental skill in various scientific and technical fields, making this exploration valuable for students and professionals alike.

At its core, a parabola is defined as the locus of points that are equidistant from a fixed point, known as the focus, and a fixed line, known as the directrix. This fundamental definition is crucial for understanding the geometry of parabolas and for deriving their equations. Let's break down this definition further: Imagine a point P(x, y) on the parabola. The distance from P to the focus must be equal to the distance from P to the directrix. The focus, in this case, is given as (-3, 0). The directrix is the vertical line x = 3. To find the equation of the parabola, we must express this geometric relationship algebraically. We accomplish this by using the distance formula to calculate the distance between the point P(x, y) and the focus (-3, 0), and the distance between the point P(x, y) and the directrix x = 3. The distance formula states that the distance between two points (x₁, y₁) and (x₂, y₂) is √[(x₂ - x₁)² + (y₂ - y₁)²]. Therefore, the distance between P(x, y) and the focus (-3, 0) is √[(x - (-3))² + (y - 0)²] which simplifies to √[(x + 3)² + y²]. The distance from a point to a line is calculated differently. For a vertical line like x = 3, the distance from a point P(x, y) is simply the absolute difference between the x-coordinate of the point and the x-value of the line. In this case, the distance from P(x, y) to the directrix x = 3 is |x - 3|. According to the definition of a parabola, these two distances must be equal. Thus, we have the equation √[(x + 3)² + y²] = |x - 3|. This equation is the algebraic representation of the geometric definition of the parabola. The next step involves simplifying this equation to obtain the standard form of a parabolic equation. This simplification process will reveal the specific characteristics of the parabola, such as its orientation and the distance between its vertex and focus.

Now that we have the fundamental equation √[(x + 3)² + y²] = |x - 3|, derived from the definition of the parabola, we need to simplify it to obtain the standard form. This involves a series of algebraic manipulations. The first step is to eliminate the square root and the absolute value. To do this, we square both sides of the equation. Squaring both sides of √[(x + 3)² + y²] = |x - 3| yields (x + 3)² + y² = (x - 3)². This step is valid because squaring both sides of an equation preserves the equality. Note that squaring the absolute value |x - 3| results in (x - 3)² because the square of any number, whether positive or negative, is positive. Next, we expand the squared terms. Expanding (x + 3)² gives x² + 6x + 9, and expanding (x - 3)² gives x² - 6x + 9. Substituting these expansions into the equation, we get x² + 6x + 9 + y² = x² - 6x + 9. Now, we can simplify the equation by canceling out terms that appear on both sides. We can subtract x² from both sides, and we can also subtract 9 from both sides. This simplifies the equation to 6x + y² = -6x. The next step is to isolate the y² term on one side of the equation. To do this, we add 6x to both sides, resulting in y² = -12x. This equation, y² = -12x, is the standard form of a parabola that opens to the left. This form immediately tells us several things about the parabola. First, because the y term is squared and the x term is not, the parabola opens either to the left or to the right. The negative sign in front of the 12x indicates that the parabola opens to the left. The coefficient of the x term, -12, is related to the distance between the vertex and the focus, which we will explore further. The derived equation, y² = -12x, is the solution to the problem. It represents the equation of the parabola with a focus at (-3, 0) and a directrix of x = 3. This process highlights the power of algebraic manipulation in translating geometric definitions into algebraic equations.

Understanding the standard forms of parabola equations is crucial for quickly identifying the key features of a parabola, such as its orientation and vertex. There are two primary standard forms, depending on whether the parabola opens horizontally or vertically. Parabolas that open horizontally have a standard form of either y² = 4ax or y² = -4ax. In these equations, 'a' represents the distance between the vertex and the focus, and also the distance between the vertex and the directrix. If the equation is in the form y² = 4ax, the parabola opens to the right. If the equation is in the form y² = -4ax, the parabola opens to the left. The vertex of these parabolas is at the origin (0, 0). Parabolas that open vertically have a standard form of either x² = 4ay or x² = -4ay. Similar to the horizontal parabolas, 'a' represents the distance between the vertex and the focus, and also the distance between the vertex and the directrix. If the equation is in the form x² = 4ay, the parabola opens upwards. If the equation is in the form x² = -4ay, the parabola opens downwards. Again, the vertex of these parabolas is at the origin (0, 0). In our specific problem, we derived the equation y² = -12x. Comparing this to the standard forms, we see that it matches the form y² = -4ax, where -4a = -12. Solving for 'a', we get a = 3. This confirms that the parabola opens to the left, and the distance between the vertex and the focus is 3 units. The vertex of this parabola is at the origin (0, 0), and the focus is at (-3, 0), as given in the problem. The directrix is the vertical line x = 3, which is also 3 units away from the vertex in the opposite direction of the focus. These standard forms provide a quick way to analyze the characteristics of a parabola once its equation is known. Furthermore, they facilitate the process of finding the equation of a parabola when given its focus, directrix, or vertex.

Having derived the equation y² = -12x, we can now confidently identify the correct option among the given choices. The original question presented four possible equations:

A. x² = -12y B. x² = 3y C. y² = 3x D. y² = -12x

By comparing our derived equation, y² = -12x, with the given options, it is clear that option D is the correct answer. This confirms our understanding of the relationship between the focus, directrix, and the equation of a parabola. Option A, x² = -12y, represents a parabola that opens downwards, which is not consistent with the given focus and directrix. Option B, x² = 3y, represents a parabola that opens upwards, also inconsistent with the given focus and directrix. Option C, y² = 3x, represents a parabola that opens to the right, which is the opposite direction of the parabola defined by the given focus and directrix. Only option D, y² = -12x, matches the equation we derived and accurately represents the parabola with a focus at (-3, 0) and a directrix of x = 3. This exercise demonstrates the importance of understanding the standard forms of parabola equations and how the coefficients relate to the parabola's orientation and dimensions. By systematically applying the definition of a parabola and using algebraic manipulation, we were able to determine the correct equation and eliminate the incorrect options. This problem-solving approach can be applied to a variety of similar problems involving conic sections and other geometric shapes. The ability to derive and analyze equations from geometric information is a valuable skill in mathematics and its applications.

In summary, we have successfully determined the equation of a parabola with a focus at (-3, 0) and a directrix of x = 3. By applying the fundamental definition of a parabola as the locus of points equidistant from the focus and directrix, we derived the equation √[(x + 3)² + y²] = |x - 3|. Through algebraic manipulation, we simplified this equation to the standard form y² = -12x. This standard form reveals that the parabola opens to the left and has a vertex at the origin. By comparing our result with the given options, we identified D. y² = -12x as the correct equation. This problem highlights the critical connection between geometric definitions and algebraic representations. The ability to translate geometric properties into algebraic equations is a cornerstone of analytic geometry and is essential for solving problems in various fields, including physics, engineering, and computer graphics. Understanding the standard forms of conic section equations, including parabolas, allows for quick identification of key features such as orientation, vertex, and focus. Furthermore, the process of deriving the equation from the definition reinforces the understanding of the underlying geometric principles. The systematic approach used in this problem—applying the definition, setting up the equation, simplifying algebraically, and comparing with standard forms—can be applied to a wide range of similar problems involving conic sections and other geometric shapes. This exploration underscores the importance of a solid foundation in both geometry and algebra for tackling mathematical problems effectively. The ability to confidently work with parabolas and other conic sections is a valuable skill for anyone pursuing further studies or a career in science, technology, engineering, or mathematics.