Finding The Foci Of An Ellipse Explained Step-by-Step

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In the fascinating world of conic sections, the ellipse holds a special place. Defined as the locus of points for which the sum of the distances to two fixed points (the foci) is constant, the ellipse exhibits a beautiful symmetry and has numerous applications in fields like astronomy, engineering, and art. This article delves into the equation of an ellipse and provides a step-by-step guide to determine the location of its foci. Our main focus will be on the equation $\frac{(x+1)^2}{225}+\frac{(y+6)^2}{144}=1$, a standard form that reveals essential information about the ellipse's geometry. We will dissect this equation, identify its key parameters, and then employ a straightforward formula to pinpoint the coordinates of the foci. By the end of this exploration, you'll not only understand how to find the foci of this particular ellipse but also gain a broader understanding of the properties and characteristics of ellipses in general. Understanding the foci is crucial for grasping the ellipse's shape and orientation, and this knowledge can be applied to various problem-solving scenarios. So, let's embark on this journey to unravel the mysteries of the ellipse and its foci.

Understanding the Ellipse Equation

To embark on our quest to find the foci of the ellipse represented by the equation $\frac{(x+1)^2}{225}+\frac{(y+6)^2}{144}=1$, we must first decode the language of the ellipse equation. The standard form of an ellipse equation centered at $(h, k)$ is given by:

$\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$

where:

• $(h, k)$ represents the coordinates of the center of the ellipse.
• $a$ is the length of the semi-major axis (the distance from the center to the farthest point on the ellipse).
• $b$ is the length of the semi-minor axis (the distance from the center to the closest point on the ellipse).

In our specific equation, $\frac{(x+1)^2}{225}+\frac{(y+6)^2}{144}=1$, we can readily identify the center and the lengths of the semi-major and semi-minor axes by comparing it to the standard form. By rewriting the equation as:

$\frac{(x-(-1))^2}{15^2} + \frac{(y-(-6))^2}{12^2} = 1$

We can directly infer the following:

• The center of the ellipse, $(h, k)$, is $(-1, -6)$.
• The semi-major axis, $a$, is $\sqrt{225} = 15$.
• The semi-minor axis, $b$, is $\sqrt{144} = 12$.

Since $a > b$, we know that the major axis is horizontal, meaning the ellipse is stretched more along the x-axis than the y-axis. This understanding is crucial for visualizing the ellipse and determining the orientation of the foci. The foci always lie on the major axis, equidistant from the center. In the next section, we will delve into the relationship between $a$, $b$, and the distance from the center to each focus, denoted by $c$. This relationship will allow us to calculate the precise location of the foci.

Calculating the Focal Distance

Having identified the center, semi-major axis, and semi-minor axis of our ellipse, the next crucial step is to determine the distance from the center to each focus. This distance, often denoted by $c$, is intimately related to the semi-major and semi-minor axes through a fundamental equation derived from the geometry of the ellipse. The relationship is expressed as:

$c^2 = a^2 - b^2$

where:

• $c$ is the distance from the center of the ellipse to each focus.
• $a$ is the length of the semi-major axis.
• $b$ is the length of the semi-minor axis.

This equation stems from the definition of an ellipse, which states that the sum of the distances from any point on the ellipse to the two foci is constant and equal to $2a$. By considering specific points on the ellipse, such as the endpoints of the minor axis, we can derive this relationship using the Pythagorean theorem. Now, let's apply this equation to our ellipse, where $a = 15$ and $b = 12$. Substituting these values into the equation, we get:

$c^2 = 15^2 - 12^2$

$c^2 = 225 - 144$

$c^2 = 81$

Taking the square root of both sides, we find:

$c = \sqrt{81} = 9$

Therefore, the distance from the center of the ellipse to each focus is 9 units. This value is critical for determining the coordinates of the foci. Since we know the center of the ellipse is $(-1, -6)$ and the major axis is horizontal, we can now move 9 units to the left and right along the major axis from the center to locate the foci. In the subsequent section, we will perform this calculation to pinpoint the precise coordinates of the foci.

Pinpointing the Foci Coordinates

With the focal distance $c = 9$ calculated, we are now poised to determine the precise coordinates of the foci. As established earlier, the foci lie on the major axis, which, in our case, is horizontal. This means that the foci will be located along the line $y = -6$, which passes through the center of the ellipse. To find the x-coordinates of the foci, we simply add and subtract the focal distance $c$ from the x-coordinate of the center. The center of the ellipse is $(-1, -6)$. Therefore, the coordinates of the foci are:

• Focus 1: $(-1 + c, -6) = (-1 + 9, -6) = (8, -6)$
• Focus 2: $(-1 - c, -6) = (-1 - 9, -6) = (-10, -6)$

Thus, the foci of the ellipse are located at the points $(8, -6)$ and $(-10, -6)$. These points are equidistant from the center, lying 9 units to the right and left, respectively. This result confirms our understanding of the ellipse's geometry and the relationship between the center, foci, and focal distance. The foci play a fundamental role in defining the shape of the ellipse. Imagine placing pins at the foci and attaching a string of length $2a$ to them. If you hold a pencil against the string and trace around the pins, you will create the ellipse. This visualization helps to solidify the concept of the foci and their importance in the ellipse's definition. In conclusion, we have successfully identified the foci of the ellipse represented by the equation $\frac{(x+1)^2}{225}+\frac{(y+6)^2}{144}=1$. The foci are located at $(8, -6)$ and $(-10, -6)$. This determination involved understanding the standard form of the ellipse equation, calculating the focal distance, and applying this distance to the center coordinates. With this knowledge, you can confidently tackle similar problems involving ellipses and their foci.

Conclusion

In this comprehensive exploration, we've successfully navigated the landscape of ellipses, focusing on the critical task of identifying the foci. Starting with the equation $\frac{(x+1)^2}{225}+\frac{(y+6)^2}{144}=1$, we meticulously dissected its components, revealing the center, semi-major axis, and semi-minor axis. We then harnessed the power of the equation $c^2 = a^2 - b^2$ to calculate the focal distance, a crucial parameter in locating the foci. Finally, armed with the focal distance and the center coordinates, we pinpointed the foci at $(8, -6)$ and $(-10, -6)$. This journey underscores the importance of understanding the standard form of conic section equations. By recognizing the key parameters embedded within these equations, we can unlock a wealth of information about the geometric properties of the curves they represent. The ellipse, with its elegant symmetry and practical applications, serves as a prime example of this principle. The ability to determine the foci is not just a mathematical exercise; it provides a deeper appreciation for the ellipse's shape and characteristics. As we've seen, the foci are not arbitrary points but rather fundamental anchors that define the curve's very essence. The process we've outlined here can be readily applied to other ellipse equations, empowering you to confidently explore the world of conic sections. Whether you're a student delving into the intricacies of mathematics or a curious mind seeking to understand the beauty of geometric forms, the knowledge gained here will serve as a valuable tool in your journey. So, embrace the power of equations, and let them guide you to new discoveries in the fascinating realm of mathematics. The answers are B. (-10,-6) and (8,-6).