Problem 14 - Algebra and Trigonometry

Question

use vertices and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes. $$ \frac{x^{2}}{16}-\frac{y^{2}}{25}=1 $$

Step-by-Step Solution

Answer

The foci are located at (± √41 , 0), and the equations of the asymptotes are y = ±5x/4.

Step 1: Calculate a and b

The equation of the hyperbola is \(\frac{x^{2}}{16} - \(\frac{y^{2}}{25} = 1 \). Thus, the square root of the coefficient of \(x^{2}\) is 4, which gives us 'a'. And the square root of the coefficient of \(y^{2}\) is 5, which gives us 'b'.

Step 2: Find the foci

The foci can be found by the equation \(c = \sqrt{a^{2} + b^{2}}\). Substituting the values of a and b, we get \(c = \sqrt{4^{2} + 5^{2}} = \sqrt{41}\), so we have two foci located at (± √41 , 0).

Step 3: Write down the equations of the asymptotes

The equations of asymptotes can be obtained by changing the 'equal to' sign to 'approximately equal to' in the general equation of the hyperbola. Therefore, the equation of the asymptotes will be \(± \frac{y}{5} = \frac{x}{4}\), or in slope-intercept form: y = ±5x/4.

Step 4: Graph the hyperbola

To graph the hyperbola: 1. Draw the x and y axes. 2. Place the center at the origin (0,0) as there are no constants in the equation. 3. Draw each vertex at a distance of 'a' units to the right and left of the center along the x-axis (±4, 0), and each co-vertex at a distance of 'b' units above and below the center along the y-axis (0, ±5). 4. Draw a dotted line through the vertices and co-vertices to represent the asymptotes. 5. Draw the hyperbola opening rightwards and leftwards passing through the foci and approaching the asymptotes as it moves away from the center.

Key Concepts

Hyperbola Asymptotes

Understanding the role of asymptotes in sketching hyperbolas is essential. Asymptotes are lines that the hyperbola approaches but never actually meets. They give us a clear outline for where the arms of a hyperbola will 'aim' as they extend infinitely. In the case of the given equation \( \frac{x^{2}}{16}-\frac{y^{2}}{25}=1 \), the asymptotes form a 'X' shape and are found by setting the hyperbola equation to zero.
For this hyperbola, the asymptotes are at \( \pm \frac{y}{5} = \frac{x}{4} \), or in slope-intercept form \( y = \pm \frac{5x}{4} \). These lines will not be crossed by the hyperbola, and they act as a guide when graphing, ensuring the curves are properly oriented within the coordinate plane.

Hyperbola Foci

The foci of a hyperbola are two points located along its major axis, which is the transverse axis in this case, and they are crucial for the hyperbola's precise definition. To find the foci of the hyperbola \( \frac{x^{2}}{16}-\frac{y^{2}}{25}=1 \), we calculate \( c \), the distance from the center to each focus, with the formula \( c = \sqrt{a^{2} + b^{2}} \).
For our equation, we have \( c = \sqrt{4^{2} + 5^{2}} = \sqrt{41} \), and since the center of this hyperbola is at the origin (0,0), the foci are at (±\sqrt{41} , 0). These points are pivotal in constructing the hyperbola's shape and ensuring it is drawn accurately.

Hyperbola Vertices

Vertices are another pair of significant points found on a hyperbola's transverse axis. They are the closest points on the hyperbola to its center and represent the 'turning points' where the curve starts to open and move away.
In our equation \( \frac{x^{2}}{16}-\frac{y^{2}}{25}=1 \), the vertices are found at a distance 'a' from the center along the x-axis. Since 'a' is 4, we plot the vertices at (±4, 0). The vertices help in defining the fundamental shape of the hyperbola and are essential for plotting its course on a graph.

Hyperbola Equation

The general form of a hyperbola equation is \( \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 \) for a horizontal hyperbola like ours, or \( \frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1 \) for a vertical one. In this equation, 'a' and 'b' are the lengths from the center to the vertices and co-vertices, respectively.
In the example \( \frac{x^{2}}{16}-\frac{y^{2}}{25}=1 \), 'a' equals 4 and 'b' equals 5. The coordinates of the center, vertices, co-vertices, foci, and the slopes of the asymptotes are all determined by these values. Understanding how these elements interconnect in the equation allows for a meaningful interpretation of the hyperbola's graph.