Problem 11 - Precalculus with Limits

Question

In Exercises 9-22, find the center, vertices, foci, and the equations of the asymptotes of the hyperbola, and sketch its graph using the asymptotes as an aid. \(\dfrac{y^2}{25}-\dfrac{x^2}{81}=1\)

Step-by-Step Solution

Answer

The center of the hyperbola is at (0,0). The vertices lie at points (0,5) and (0,-5). The foci are at points (0, \(\sqrt{106}\)) and (0, -\(\sqrt{106}\)). The equations for the asymptotes are \(y=\frac{5x}{9}\) and \(y=-\frac{5x}{9}\).

Step 1: Find the center

The center of the hyperbola is given by the point \((h, k)\). Here, since there are no values added or subtracted from \(x\) and \(y\) in the equation's denominators, the center is at the origin which is \((0,0)\).

Step 2: Calculate the vertices

The vertices of the hyperbola along the y-axis are given by \((h, k\pm a)\). Here, since \(a^2 = 25\), \(a = 5\). Therefore, the vertices are located at \((0, 0\pm 5)\) which translates to two points \((0, 5)\) and \((0, -5)\).

Step 3: Find the foci

The foci fall along the y-axis and they are given by \((h, k\pm c)\), where \(c =\sqrt{a^2 + b^2} = \sqrt{25+81} = \sqrt{106}\). Hence, the foci are at \((0, 0\pm \sqrt{106})\) which corresponds to the points \((0, \sqrt{106})\) and \((0, -\sqrt{106})\).

Step 4: Find the equations of the asymptotes

The equations of the asymptotes for this type of hyperbola are given by \(y = k\pm \frac{a}{b}(x-h)\). Plugging in our values, we get \(y =\pm \frac{5}{9}x\). Hence, the equations of the asymptotes are \(y=\frac{5x}{9}\) and \(y=-\frac{5x}{9}\).

Step 5: Sketch the graph

The graph should be sketched using the center, vertices, foci, and asymptotes as aids. Since these are hard to portray in text format, refer to any number of online graphing calculators or tools available in textbooks to get a visual representation.

Key Concepts

Asymptotes

Asymptotes are lines that a curve approaches as it extends towards infinity. For hyperbolas in coordinate geometry, they play a crucial role in sketching the graph accurately. They are not part of the hyperbola but act as guides to show how the hyperbola behaves at its edges.

• To find the asymptotes for a hyperbola of the form \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), we use the formula: \(y = k \pm \frac{a}{b}(x-h)\).
• In our exercise, the center \((h, k)\) is at the origin \((0,0)\), making the equation simplify to \(y = \pm \frac{a}{b}x\).
• With \(a = 5\) and \(b = 9\), the asymptotes are: \(y = \frac{5}{9}x\) and \(y = -\frac{5}{9}x\).

These lines give you a structure within which the hyperbola's branches open, guiding you towards drawing it accurately on a coordinate plane.

Vertices

Vertices of a hyperbola are the points where the hyperbola curves turn around on their closest or farthest paths from the center. They signify the point where each branch of the hyperbola is closest to or farthest along its transverse axis.

• For the hyperbola \(\frac{y^2}{25} - \frac{x^2}{81} = 1\), the vertices are given by the formula \((h, k \pm a)\). Here \(a\) is found from \(a^2 = 25\), hence \(a = 5\).
• Since \((h, k)\) is at the origin \((0,0)\), the vertices are \((0, 5)\) and \((0, -5)\).

Vertices are fundamental to understanding the size and position of the hyperbola on a graph, giving you a concrete starting point for sketching its shape.

Foci

The foci (singular: focus) of a hyperbola are two fixed points that are located inside each curve. They are used to define the hyperbola and play a critical role in its geometric properties.

• In a hyperbola described by \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), the foci are found using \((h, k \pm c)\), where \(c = \sqrt{a^2 + b^2}\).
• For our specific equation, \(a^2 = 25\) and \(b^2 = 81\), leading to \(c = \sqrt{106}\).
• This places the foci at \((0, \sqrt{106})\) and \((0, -\sqrt{106})\).

Foci help in understanding how a hyperbola stretches and provide insight into its reflective properties and dimensions.

Coordinate Geometry

Coordinate geometry, or analytic geometry, is a mathematical discipline that uses algebra to study geometric problems. It provides the framework for plotting hyperbolas on a coordinate plane.

• In coordinate geometry, we analyze the position and equations of shapes, like hyperbolas, using coordinates often labeled as \((x, y)\).
• Let’s say for a hyperbola such as \(\frac{y^2}{25} - \frac{x^2}{81} = 1\), coordinate geometry allows us to pinpoint the exact places for its center, vertices, foci, and asymptotes.
• This approach makes solving geometric problems more systematic, as you can visualize and plot elements on a two-dimensional graph.

Through this system, even complex shapes like hyperbolas can be understood and sketched accurately, giving life to the geometric language of mathematics.