Finding The Equation Of A Parabola With Focus (-1, 15) And Directrix X = -4

In the realm of conic sections, the parabola stands out as a fascinating curve with a unique definition and a wide array of applications. A parabola is defined as the set of all points in a plane that are equidistant from a fixed point, called the focus, and a fixed line, called the directrix. Understanding this definition is crucial to deriving the equation of a parabola when given its focus and directrix. This article will guide you through the process of finding the equation of a parabola, focusing on a specific example where the focus is at $(-1, 15)$ and the directrix is the line $x = -4$. We will explore the fundamental properties of parabolas, the standard forms of their equations, and the algebraic manipulations required to arrive at the correct answer. By the end of this guide, you will have a solid understanding of how to determine the equation of a parabola from its defining characteristics.

Before we delve into the specific problem, it's essential to grasp the fundamental concepts of a parabola. As mentioned earlier, a parabola is defined by its focus and directrix. The focus is a fixed point, and the directrix is a fixed line. The vertex, a crucial point on the parabola, is the point that lies exactly halfway between the focus and the directrix. The axis of symmetry is the line that passes through the focus and the vertex, perpendicular to the directrix. The distance between the vertex and the focus (or the vertex and the directrix) is commonly denoted by 'p'. This parameter 'p' plays a significant role in the equation of the parabola. The sign of 'p' determines the direction in which the parabola opens. If p > 0, the parabola opens to the right or upwards, depending on the orientation of the directrix. If p < 0, the parabola opens to the left or downwards. In our case, the focus is $(-1, 15)$ and the directrix is $x = -4$. This setup implies that the parabola will open to the right, as the focus is to the right of the directrix. Understanding these basics will help us to construct the equation of the parabola in the subsequent sections.

To find the equation of the parabola, we need to use the definition that any point $(x, y)$ on the parabola is equidistant from the focus and the directrix. Let's denote the focus as $F(-1, 15)$ and consider a general point $P(x, y)$ on the parabola. The distance between $P$ and $F$, denoted as $PF$, can be calculated using the distance formula: $PF = \sqrt{(x - (-1))^2 + (y - 15)^2} = \sqrt{(x + 1)^2 + (y - 15)^2}$. The distance from the point $P(x, y)$ to the directrix $x = -4$ is the perpendicular distance, which is simply the absolute difference in the x-coordinates: $|x - (-4)| = |x + 4|$. According to the definition of a parabola, these two distances must be equal: $PF = |x + 4|$. Thus, we have the equation: $\sqrt{(x + 1)^2 + (y - 15)^2} = |x + 4|$. To eliminate the square root, we square both sides of the equation: $(x + 1)^2 + (y - 15)^2 = (x + 4)^2$. Now, we expand the squared terms: $x^2 + 2x + 1 + (y - 15)^2 = x^2 + 8x + 16$. We can simplify this equation by canceling the $x^2$ terms and rearranging: $(y - 15)^2 = 6x + 15$. To isolate $x$, we divide both sides by 6: $x = \frac{1}{6}(y - 15)^2 - \frac{5}{2}$. This equation represents the parabola with the given focus and directrix. Comparing this result with the provided options, we find that it matches option A.

Let's recap the steps we took to arrive at the solution. First, we defined the parabola in terms of its focus and directrix. We then expressed the distance between a general point $(x, y)$ on the parabola and the focus $(-1, 15)$, and the distance between the same point and the directrix $x = -4$. By equating these distances, we obtained the fundamental equation: $\sqrt{(x + 1)^2 + (y - 15)^2} = |x + 4|$. Squaring both sides to eliminate the square root, we got: $(x + 1)^2 + (y - 15)^2 = (x + 4)^2$. Expanding the terms yielded: $x^2 + 2x + 1 + (y - 15)^2 = x^2 + 8x + 16$. Simplifying the equation by canceling the $x^2$ terms and rearranging, we arrived at: $(y - 15)^2 = 6x + 15$. Finally, we isolated $x$ to obtain the equation in the desired form: $x = \frac{1}{6}(y - 15)^2 - \frac{15}{6}$, which simplifies to $x = \frac{1}{6}(y - 15)^2 - \frac{5}{2}$. This step-by-step approach not only confirms the correctness of the solution but also provides a clear understanding of the underlying principles and algebraic manipulations involved. Each step is logically connected, making it easier to follow and replicate the process for similar problems. The final equation matches option A, which is the correct answer. This detailed solution demonstrates how the definition of a parabola can be directly applied to derive its equation, emphasizing the importance of understanding the relationship between the focus, directrix, and the points on the curve.

In conclusion, determining the equation of a parabola given its focus and directrix involves a clear understanding of the parabola's definition and the application of the distance formula. By equating the distances from a general point on the parabola to the focus and the directrix, we can derive the equation that represents the parabola. The algebraic manipulations, including squaring, expanding, and simplifying, are crucial steps in arriving at the final form. The step-by-step approach outlined in this article provides a systematic way to solve such problems, ensuring accuracy and clarity. Understanding the relationship between the focus, directrix, vertex, and the parameter 'p' is essential for mastering parabola equations. Through practice and a solid grasp of the fundamental concepts, you can confidently tackle a wide range of problems involving parabolas. The ability to derive and manipulate parabola equations is not only valuable in mathematics but also has applications in physics, engineering, and other fields. This article has provided a comprehensive guide to finding the equation of a parabola, equipping you with the knowledge and skills to excel in this area of mathematics.

The correct answer is A. $x = \frac{1}{6}(y - 15)^2 - \frac{5}{2}$.