Hyperbolas at the Origin: Videos & Practice Problems

Hyperbolas are unique conic sections characterized by their equations, which differ from ellipses by a minus sign. The vertices are located at a distance a2 from the center, while the foci are determined using c^2 = a^2 + b^2. Asymptotes are essential for graphing, formed by drawing lines through the vertices and b points. Understanding these concepts is crucial for accurately graphing hyperbolas and identifying their properties.

Introduction to Hyperbolas

In the study of conic sections, hyperbolas present a unique and often challenging shape to understand. The equation of a hyperbola closely resembles that of an ellipse, with the key distinction being the presence of a minus sign instead of a plus sign. For a horizontal hyperbola, the general equation can be expressed as:

\[\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\]

Here, the value of a represents the distance from the center of the hyperbola to each of its two branches, while b plays a crucial role in determining the shape and orientation of the hyperbola, particularly when graphing. The visual representation of a hyperbola consists of two curves that open away from each other, resembling two parabolas.

For a vertical hyperbola, the equation takes the form:

\[\frac{x^2}{b^2} - \frac{y^2}{a^2} = 1\]

In this case, a still indicates the distance from the center to the curves, but the orientation is now vertical. The b value, while less intuitive than in ellipses, is essential for graphing and understanding the hyperbola's dimensions.

It is important to note that in hyperbolas, the a value does not necessarily represent the largest number in the equation; rather, it is the first term in the equation. This contrasts with ellipses, where a is always the largest value.

To effectively graph hyperbolas, one must identify the orientation based on the position of the squared terms. For instance, if y is squared first, the hyperbola opens vertically, while if x is squared first, it opens horizontally. Understanding these relationships allows for accurate graphing and interpretation of hyperbolas.

In practice, when given an equation, one can determine the corresponding graph by identifying the values of a and b, and then plotting the curves based on their respective distances from the center. This method reinforces the connection between the algebraic representation of hyperbolas and their geometric forms.

Given the hyperbola $\frac{x^2}{25}-\frac{y^2}{9}=1$, find the length of the $a$-axis and the $b$ -axis.

• A. $a=25$, $b=9$
• B. $a=9,b=25$
• C. $a=5,b=3$
• D. $a=3,b=5$

Answer: (C)

Given the hyperbola $x^2-\frac{y^2}{4}=1$, find the length of the $a$ -axis and $b$ -axis.

• A. $a=1$, $b=4$
• B. $a=4$ , $b=1$
• C. $a=1$, $b=2$
• D. $a=2$, $b=1$

Answer: (C)

Given the hyperbola $\frac{y^2}{100}-\frac{x^2}{139}=1$, find the length of the $a$ -axis and the $b$ -axis.

• A. $a=100$, $b=139$
• B. $a=139$, $b=100$
• C. $a=\sqrt{139}$, $b=10$
• D. $a=10$, $b=\sqrt{139}$

Answer: (D)

Foci and Vertices of Hyperbolas

In the study of hyperbolas, understanding how to find the vertices and foci is crucial. A hyperbola, like an ellipse, has two vertices and two foci located along its major axis. The vertices are the points closest to the center of the hyperbola, and the distance from the center to each vertex is denoted as a.

To find the vertices, one must first identify the center of the hyperbola, which is often given in the standard form of the equation. For a hyperbola expressed in the form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the value of \(a^2\) is the first term in the denominator. For example, if \(a^2 = 4\), then \(a = \sqrt{4} = 2\). Starting from the center, you can locate the vertices by moving a units to the left and right along the x-axis, resulting in vertices at \((-2, 0)\) and \((2, 0)\) for a hyperbola centered at the origin.

The foci of a hyperbola are unique points where the difference in distances from any point on the hyperbola to each focus remains constant. To calculate the foci, you need to determine the distance c, which is derived from the equation \(c^2 = a^2 + b^2\). Here, \(b^2\) is the second term in the denominator of the hyperbola's standard form. For instance, if \(b^2 = 9\), then \(b = \sqrt{9} = 3\). Using the earlier example where \(a^2 = 4\), we find \(c^2 = 4 + 9 = 13\), leading to \(c = \sqrt{13} \approx 3.6\). The foci are then located at \((\sqrt{13}, 0)\) and \((- \sqrt{13}, 0)\).

It is important to note the orientation of the hyperbola affects the placement of the vertices and foci. For a horizontal hyperbola, both the vertices and foci lie on the x-axis, while for a vertical hyperbola, they are positioned on the y-axis. In the case of a vertical hyperbola, the vertices would be at \((0, a)\) and \((0, -a)\), and the foci would be at \((0, c)\) and \((0, -c)\).

In summary, finding the vertices and foci of a hyperbola involves understanding the relationship between the parameters \(a\), \(b\), and \(c\), and recognizing how the hyperbola's orientation influences their positions. Mastery of these concepts is essential for solving problems related to hyperbolas in mathematics.

Determine the vertices and foci of the hyperbola $\frac{y^2}{4}-x^2=1$.

• A. Vertices: $\left(2,0\right),\left(-2,0\right)$Foci: $\left(\sqrt5,0\right),\left(-\sqrt5,0\right)$
• B. Vertices: $\left(0,2\right),\left(0,-2\right)$Foci: $\left(0,\sqrt5\right),\left(0,-\sqrt5\right)$
• C. Vertices: $\left(1,0\right),\left(-1,0\right)$Foci: $\left(5,0\right),\left(-5,0\right)$
• D. Vertices: $\left(0,1\right),\left(0,-1\right)$Foci: $\left(0,5\right),\left(0,-5\right)$

Answer: (B)

Find the equation for a hyperbola with a center at $\left(0,0\right)$, focus at $\left(0,-6\right)$ and vertex at $\left(0,4\right)$.

• A. $\frac{y^2}{16}-\frac{x^2}{20}=1$
• B. $\frac{y^2}{20}-\times\frac{^2}{16}=1$
• C. $\frac{y^2}{4}-\frac{x^2}{\sqrt{20}}=1$
• D. $\frac{y^2}{\sqrt{20}}-\frac{x2}{4}=1$

Answer: (A)

Asymptotes of Hyperbolas

Understanding the asymptotes of hyperbolas is crucial for graphing these shapes accurately. Asymptotes are lines that help define the behavior of hyperbolas as they extend towards infinity. To find the asymptotes, we utilize the values of a and b, which represent the distances from the center to the vertices and co-vertices, respectively. For a hyperbola in standard form, the relationship is given by a² and b² in the denominators of the equation.

To illustrate, consider a vertical hyperbola where a² = 9 and b² = 4. From these, we find a = 3 and b = 2. These values help us construct a rectangle (or box) that connects the vertices and co-vertices. The asymptotes can then be drawn through the corners of this box, extending through the center of the hyperbola.

The equations of the asymptotes for a vertical hyperbola can be derived from the slope formula, where the slope m is calculated as the rise over the run. In this case, the rise corresponds to a and the run corresponds to b. Thus, the equations for the asymptotes are:

y = ±(a/b)x

For our example, this results in:

y = ±(3/2)x

When dealing with a horizontal hyperbola, the process is similar, but the roles of a and b are switched. The equations for the asymptotes become:

y = ±(b/a)x

Using the same values, for a horizontal hyperbola with a = 3 and b = 2, the asymptote equations would be:

y = ±(2/3)x

This systematic approach allows for quick determination of asymptotes, enhancing the efficiency of graphing hyperbolas. By recognizing the patterns in the relationships between a and b, students can confidently find the asymptotes for both vertical and horizontal hyperbolas.

Find the equations for the asymptotes of the hyperbola $\frac{x^2}{64}-\frac{y^2}{100}=1$.

• A. $y=\pm\frac45x$
• B. $y=\pm\frac54x$
• C. $y=\pm\frac{16}{25}x$
• D. $y=\pm\frac{25}{16}x$

Answer: (B)

Find the equations for the asymptotes of the hyperbola $\frac{y^2}{16}-\frac{x^2}{9}=1$.

• A. $y=\pm\frac{9}{16}x$
• B. $y=\pm\frac{16}{9}x$
• C. $y=\pm\frac34x$
• D. $y=\pm\frac43x$

Answer: (D)

Graph Hyperbolas at the Origin

To graph a hyperbola from its equation, it is essential to first determine its orientation—whether it is horizontal or vertical. This can be identified by examining the equation; if the term with \(x^2\) appears first in the denominator, the hyperbola is horizontal. For example, if the equation is of the form \(\frac{(y-k)^2}{b^2} - \frac{(x-h)^2}{a^2} = 1\), it indicates a vertical hyperbola, while \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\) indicates a horizontal hyperbola.

Next, identify the vertices of the hyperbola. For a horizontal hyperbola, the vertices are located at \((h \pm a, k)\), where \(a\) is derived from \(a^2\) in the equation. For instance, if \(a^2 = 9\), then \(a = 3\), placing the vertices at \((h+3, k)\) and \((h-3, k)\).

Following this, determine the \(b\) values, which are found from \(b^2\) in the equation. If \(b^2 = 64\), then \(b = 8\). The \(b\) points for a horizontal hyperbola are located at \((h, k \pm b)\), resulting in points at \((h, k+8)\) and \((h, k-8)\).

To find the asymptotes, draw a rectangle using the vertices and \(b\) points. The asymptotes are then represented by lines that pass through the corners of this rectangle, following the equations \(y - k = \pm \frac{b}{a}(x - h)\).

Finally, sketch the branches of the hyperbola, which approach the asymptotes but never intersect them. The general shape of the hyperbola will open towards the direction of the vertices.

To locate the foci, use the relationship \(c^2 = a^2 + b^2\). For example, if \(a^2 = 9\) and \(b^2 = 64\), then \(c^2 = 9 + 64 = 73\), leading to \(c = \sqrt{73} \approx 8.54\). The foci for a horizontal hyperbola are positioned at \((h \pm c, k)\), which means they will be at approximately \((h+8.54, k)\) and \((h-8.54, k)\).

By following these steps, one can effectively graph a hyperbola and identify its key features, including vertices, asymptotes, and foci, all derived from the equation.

Graphing Hyperbolas at the Origin Example 1

In this example, we explore how to graph a hyperbola, focusing on its orientation, vertices, b points, asymptotes, and foci. The first step is to determine the orientation of the hyperbola by examining the equation. Since the term with \(x^2\) appears first, we conclude that the hyperbola is oriented horizontally along the x-axis.

Next, we identify the vertices. The value of \(a^2\) is derived from the first denominator, which is 16. Taking the square root gives us \(a = \sqrt{16} = 4\). Therefore, the vertices are located at the points (4, 0) and (-4, 0) on the graph.

Following this, we calculate the b points, which are positioned vertically opposite the vertices. Here, \(b^2\) corresponds to the second number in the denominator, which is 20. Thus, \(b = \sqrt{20}\), which can be simplified to \(2\sqrt{5}\) or approximately 4.47. The b points are then at (0, \( \sqrt{20} \)) and (0, \(-\sqrt{20}\)), or approximately (0, 4.47) and (0, -4.47).

To find the asymptotes, we draw a rectangle (or box) that connects the vertices and b points. The asymptotes are represented by lines that extend through the corners of this box. These lines help define the behavior of the hyperbola as it approaches these asymptotes.

Next, we sketch the branches of the hyperbola, which start at the vertices and approach the asymptotes. One branch extends from (4, 0) towards the upper asymptote and downwards, while the other branch starts at (-4, 0) and approaches the lower asymptote.

Finally, we determine the foci of the hyperbola. For a horizontal hyperbola, the foci are located along the x-axis. We use the relationship \(c^2 = a^2 + b^2\) to find \(c\). Here, \(a^2 = 16\) and \(b^2 = 20\), leading to \(c^2 = 16 + 20 = 36\). Taking the square root gives us \(c = \sqrt{36} = 6\). Thus, the foci are positioned at (6, 0) and (-6, 0).

In summary, we have successfully graphed the hyperbola, identified its vertices at (4, 0) and (-4, 0), b points at (0, \( \sqrt{20} \)) and (0, \(-\sqrt{20}\)), drawn the asymptotes, and located the foci at (6, 0) and (-6, 0).

Here’s what students ask on this topic:

What is the standard form equation of a hyperbola centered at the origin?

The standard form equation of a hyperbola centered at the origin can be written in two forms, depending on its orientation:

1. For a horizontal hyperbola: $\frac{x^2}{a^2}-\frac{y^2}{b^2}=\; 1$

2. For a vertical hyperbola: $\frac{y^2}{a^2}-\frac{x^2}{b^2}=\; 1$

In these equations, $a^2$ and $b^2$ are the squares of the distances from the center to the vertices and co-vertices, respectively.

How do you find the vertices and foci of a hyperbola?

To find the vertices and foci of a hyperbola, follow these steps:

1. Identify the values of $a^2$ and $b^2$ from the standard form equation.

2. Calculate $a$ and $b$ by taking the square roots of $a^2$ and $b^2$.

3. For a horizontal hyperbola, the vertices are at $(\pm a,0)$. For a vertical hyperbola, the vertices are at $(0,\pm a)$.

4. Calculate the distance to the foci using $c^2=a^2+b^2$. Then, find $c$ by taking the square root of $c^2$.

5. For a horizontal hyperbola, the foci are at $(\pm c,0)$. For a vertical hyperbola, the foci are at $(0,\pm c)$.

How do you find the asymptotes of a hyperbola?

To find the asymptotes of a hyperbola, follow these steps:

1. Identify the values of $a^2$ and $b^2$ from the standard form equation.

2. Calculate $a$ and $b$ by taking the square roots of $a^2$ and $b^2$.

3. For a horizontal hyperbola, the equations of the asymptotes are $y=\pm \frac{b}{a}x$.

4. For a vertical hyperbola, the equations of the asymptotes are $y=\pm \frac{a}{b}x$.

5. Draw the asymptotes by plotting the lines through the center of the hyperbola, using the slopes calculated from the equations above.

How do you graph a hyperbola given its equation?

To graph a hyperbola given its equation, follow these steps:

1. Determine if the hyperbola is horizontal or vertical by looking at the standard form equation. If $\frac{x^2}{a^2}$ comes first, it is horizontal. If $\frac{y^2}{a^2}$ comes first, it is vertical.

2. Identify the vertices by calculating $a$ from $a^2$. For a horizontal hyperbola, vertices are at $(\pm a,0)$. For a vertical hyperbola, vertices are at $(0,\pm a)$.

3. Identify the co-vertices by calculating $b$ from $b^2$. For a horizontal hyperbola, co-vertices are at $(0,\pm b)$. For a vertical hyperbola, co-vertices are at $(\pm b,0)$.

4. Draw a rectangle using the vertices and co-vertices.

5. Draw the asymptotes by connecting the diagonals of the rectangle.

6. Sketch the hyperbola by drawing curves that approach the asymptotes and pass through the vertices.

What is the relationship between the hyperbola and its asymptotes?

The asymptotes of a hyperbola are lines that the hyperbola approaches but never intersects. They provide a framework for graphing the hyperbola and help define its shape. The equations of the asymptotes are derived from the values of $a$ and $b$ in the standard form equation of the hyperbola. For a horizontal hyperbola, the asymptotes are $y=\pm \frac{b}{a}x$, and for a vertical hyperbola, they are $y=\pm \frac{a}{b}x$. The hyperbola's branches get closer to the asymptotes as they extend further from the center, but they never touch the asymptotes.