Find Ellipse Vertices: X²/121 + Y²/1 = 1

Understanding the vertices of an ellipse is crucial for grasping its geometry and characteristics. The equation $\frac{x^2}{121}+\frac{y^2}{1}=1$ represents a specific ellipse, and identifying its vertices helps us pinpoint the farthest points along its major axis. In this article, we'll break down how to find these key points, explore the concepts behind them, and determine which of the given options correctly identifies the vertices for this particular ellipse. We'll delve into the standard form of an ellipse equation and how the denominators (the numbers beneath the $x^2$ and $y^2$ terms) dictate the shape and orientation of the ellipse. The larger denominator, when associated with a variable, tells us the direction of the major axis. In our equation, 121 is under $x^2$ and 1 is under $y^2$. Since 121 is significantly larger than 1, the major axis lies along the x-axis. The vertices are the endpoints of this major axis. To find them, we take the square root of the larger denominator and place it with a plus and minus sign on the axis corresponding to the larger denominator. So, for $\frac{x^2}{121}+\frac{y^2}{1}=1$, the larger denominator is 121, which is under $x^2$. The square root of 121 is 11. Therefore, the vertices will be located at $(\pm 11, 0)$. This means the points are $(-11,0)$ and $(11,0)$. These are the points that are furthest apart on the ellipse, lying along its longest diameter. It's important to remember that the vertices are distinct from the co-vertices, which lie at the endpoints of the minor axis. The co-vertices are found by taking the square root of the smaller denominator and placing it with a plus and minus sign on the axis corresponding to the smaller denominator. In this case, the smaller denominator is 1, and its square root is 1. So, the co-vertices would be at $(0, \pm 1)$, or $(0,-1)$ and $(0,1)$. By distinguishing between vertices and co-vertices, we gain a more complete understanding of the ellipse's dimensions and shape. The equation of an ellipse is typically presented in the standard form: $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$ for an ellipse centered at $(h, k)$. In our case, the ellipse is centered at the origin $(0,0)$, so $h=0$ and $k=0$. The values of $a^2$ and $b^2$ are the denominators. When $a^2 > b^2$, the major axis is horizontal, and the vertices are at $(h \pm a, k)$. When $b^2 > a^2$, the major axis is vertical, and the vertices are at $(h, k \pm b)$. In the equation $\frac{x^2}{121}+\frac{y^2}{1}=1$, we have $a^2 = 121$ and $b^2 = 1$. Since $121 > 1$, we know that $a^2$ is associated with the $x^2$ term, indicating a horizontal major axis. The value of $a$ is the square root of $a^2$, so $a = \sqrt{121} = 11$. Since the center is $(0,0)$, the vertices are at $(0 \pm 11, 0)$, which simplifies to $(-11, 0)$ and $(11, 0)$. These points represent the extreme horizontal extent of the ellipse. If the equation were, for example, $\frac{x^2}{1}+\frac{y^2}{121}=1$, then $b^2 = 121$ and $a^2 = 1$. In that scenario, $b^2 > a^2$, meaning the major axis would be vertical, and the vertices would be at $(0, \pm 121^{1/2})$, which are $(0, -11)$ and $(0, 11)$. This distinction is critical for correctly identifying the vertices. The numbers under the squared terms, $a^2$ and $b^2$, are fundamental to understanding the ellipse's dimensions. The larger of these two numbers determines the length of the semi-major axis, and its position under $x^2$ or $y^2$ determines the orientation of the major axis. The square root of this larger number gives us the distance from the center to each vertex along the major axis. Conversely, the square root of the smaller number gives us the length of the semi-minor axis, and its position determines the orientation of the minor axis. The endpoints of the minor axis are called co-vertices. For our specific equation, $\frac{x^2}{121}+\frac{y^2}{1}=1$, the larger denominator is 121, found under the $x^2$ term. This immediately tells us that the major axis is horizontal and lies along the x-axis. The value of the semi-major axis, denoted by 'a', is the square root of 121, which is 11. Since the ellipse is centered at the origin (0,0), the vertices are located at a distance of 'a' units to the left and right of the center. Thus, the vertices are at $(0 - 11, 0)$ and $(0 + 11, 0)$, which are $(-11, 0)$ and $(11, 0)$. These are the points where the ellipse intersects its major axis, representing its widest extent horizontally. It's essential to differentiate these from the co-vertices, which lie on the minor axis. The minor axis here is vertical, and its length is determined by the square root of the smaller denominator, which is $\sqrt{1} = 1$. Therefore, the co-vertices are at $(0, 0 - 1)$ and $(0, 0 + 1)$, namely $(0, -1)$ and $(0, 1)$. These points represent the ellipse's narrowest extent vertically. The precise identification of vertices is directly tied to the comparison of the denominators in the standard form of the ellipse equation. The larger denominator indicates the direction of the major axis, and its square root gives the distance from the center to the vertices. By carefully examining the given equation $\frac{x^2}{121}+\frac{y^2}{1}=1$, we can confidently determine that the vertices are at $(-11, 0)$ and $(11, 0)$. This understanding allows us to visualize the ellipse accurately and solve related problems concerning its properties.

Understanding the Equation of an Ellipse

The standard form of an ellipse centered at the origin $(0,0)$ is given by the equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Here, $a^2$ and $b^2$ are the squares of the lengths of the semi-axes. The values of $a$ and $b$ represent the distances from the center to the endpoints of the major and minor axes, respectively. The orientation of the ellipse—whether it's wider than it is tall, or taller than it is wide—is determined by which denominator is larger. If $a^2 > b^2$, the major axis is horizontal (along the x-axis), and its length is $2a$. The vertices, which are the endpoints of the major axis, are located at $(\pm a, 0)$. If $b^2 > a^2$, the major axis is vertical (along the y-axis), and its length is $2b$. The vertices are located at $(0, \pm b)$. In contrast, the minor axis is always perpendicular to the major axis, and its length is $2 \min(a, b)$. The endpoints of the minor axis are called co-vertices.

For the specific equation provided, $\frac{x^2}{121}+\frac{y^2}{1}=1$, we can identify the values of $a^2$ and $b^2$. We have $a^2 = 121$ and $b^2 = 1$. Since $121 > 1$, it means that $a^2$ is the larger denominator and is associated with the $x^2$ term. This confirms that the major axis is horizontal and lies along the x-axis.

The length of the semi-major axis, $a$, is the square root of $a^2$. Therefore, $a = \sqrt{121} = 11$. Since the ellipse is centered at the origin $(0,0)$, the vertices are located at a distance of $a$ units to the left and right of the center. This gives us the coordinates of the vertices as $(-a, 0)$ and $(a, 0)$, which are $(-11, 0)$ and $(11, 0)$. These points represent the farthest extent of the ellipse along its horizontal dimension.

It's important to distinguish the vertices from the co-vertices. The co-vertices lie at the endpoints of the minor axis. The length of the semi-minor axis, $b$, is the square root of $b^2$. So, $b = \sqrt{1} = 1$. Since the minor axis is vertical in this case, the co-vertices are located at $(0, -b)$ and $(0, b)$, which are $(0, -1)$ and $(0, 1)$. These points represent the farthest extent of the ellipse along its vertical dimension.

By carefully analyzing the denominators in the standard equation of an ellipse, we can determine its orientation, the lengths of its major and minor axes, and consequently, the coordinates of its vertices and co-vertices. For $\frac{x^2}{121}+\frac{y^2}{1}=1$, the vertices are definitively at $(-11, 0)$ and $(11, 0)$. This systematic approach ensures accuracy when working with ellipses and their properties.

Identifying the Vertices from the Given Options

Now that we have determined the vertices of the ellipse represented by the equation $\frac{x^2}{121}+\frac{y^2}{1}=1$ to be $(-11, 0)$ and $(11, 0)$, let's examine the provided options:

A. $(0,-11)$ and $(0,11)$: These points would be the vertices if the major axis were vertical and $b^2 = 121$. This is not the case here.

B. $(0,-1)$ and $(0,1)$: These points are the co-vertices of the ellipse, as they correspond to the endpoints of the minor axis.

C. $(-11,0)$ and $(11,0)$: These points match our calculated vertices. They lie on the x-axis, which is the major axis in this instance, at a distance of 11 units from the origin.

D. $(-1,0)$ and $(1,0)$: These points would be the co-vertices if the major axis were horizontal and $a^2 = 1$. This is not the case here.

Therefore, the correct option representing the vertices of the ellipse $\frac{x^2}{121}+\frac{y^2}{1}=1$ is C.

Conclusion: Mastering Ellipse Vertices

In conclusion, finding the vertices of an ellipse is a straightforward process once you understand the standard form of its equation and the significance of the denominators. For the ellipse defined by $\frac{x^2}{121}+\frac{y^2}{1}=1$, we identified that the larger denominator, 121, is under the $x^2$ term. This tells us that the major axis is horizontal. The square root of 121 is 11, which represents the distance from the center to the vertices. Since the ellipse is centered at the origin, the vertices are located at $(-11,0)$ and $(11,0)$. This corresponds directly to option C. By understanding these principles, you can confidently tackle any problem involving ellipse vertices and their properties. Remember to always compare the denominators to determine the orientation of the major axis and then calculate the square roots to find the lengths of the semi-major and semi-minor axes.

For further exploration into conic sections and their properties, you can visit Wolfram MathWorld or Khan Academy's Conic Sections section.