Problem 22 - College Algebra

Question

Use vertices and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes. $$4 x^{2}-25 y^{2}=100$$

Step-by-Step Solution

Answer

The graph of the hyperbola will be horizontally oriented since the x-term is positive in the equation. The vertices are at (5, 0) and (-5, 0), the asymptotes are the lines \(y = \pm\frac{2}{5}x\), and the foci reside at approximately (5.39, 0) function and (-5.39, 0).

Step 1: Rewrite the equation

First, rewrite the given equation in standard form for a hyperbola. The standard form is \(\frac{(x-h)^{2}}{a^{2}} - \frac{(y-k)^{2}}{b^{2}} = 1\), where (h, k) is the center of the hyperbola, and a and b are related to the lengths of the axes (major and minor). In the given equation, this can be done by dividing all terms by 100, which will bring it to the form \(\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}}=1\).

Step 2: Identify a and b

To identify the values of a and b, take their square roots from the denominators in the equation. Here, \(a^2 = 25\) and \(b^2 = 4\), so we can find that \(a = 5\) and \(b = 2\). These values will be used in the next steps.

Step 3: Find the vertices

Because our x-term is positive in our equation, the center of our hyperbola is at the origin (0, 0). Therefore, the vertices are \((a, 0)\) and \((-a, 0)\), or in this case, (5, 0) and (-5, 0). These are the intersecting points of the hyperbola with its major axis, which in this case, lies on the x-axis.

Step 4: Find the equation of the asymptotes

The equations of the asymptotes of a hyperbola are given by \(y = \pm\frac{b}{a}x\), which means in this case it will be \(y = \pm\frac{2}{5}x\). The asymptotes form a boundary that the hyperbola's curve approaches but never crosses.

Step 5: Find the foci

The foci are given by the equation \((h \pm c, k)\) for a horizontal major axis, where c is the distance from the center to a focus, and can be calculated by \(c = \sqrt{a^2 + b^2}\). Substituting the values in, we get \(c = \sqrt{25 + 4} = \sqrt{29}\). Therefore, the coordinates of the foci will be (0 \pm \sqrt{29}, 0), or approximately (5.39, 0) and (-5.39, 0).

Key Concepts

Graphing Conic Sections

A hyperbola is one of the four main types of conic sections, which also include circles, ellipses, and parabolas. Think of a hyperbola as a type of open curve. It consists of two separate but symmetrical branches. The general equation for a hyperbola can vary depending on its orientation, but in general, it involves a subtraction between two squared terms. This equation takes the form of

• \( \frac{(x-h)^{2}}{a^{2}} - \frac{(y-k)^{2}}{b^{2}} = 1 \) for hyperbolas opening along the x-axis, or
• \( \frac{(y-k)^{2}}{b^{2}} - \frac{(x-h)^{2}}{a^{2}} = 1 \) for those opening along the y-axis.

To graph these, identify key elements like the center \((h, k)\), vertices, and asymptotes. The graph of a hyperbola tends to get closer to the lines represented by its asymptotes but never actually touches them. This visual representation helps in understanding the nature of hyperbolas.

Equations of Asymptotes

Asymptotes are crucial lines for hyperbolas, as they provide the directional boundary the hyperbola's branches follow. For a hyperbola with a horizontal transverse axis, the equations of the asymptotes can be determined using the formula:

• \( y = \pm \frac{b}{a}x \)

In our specific case for the hyperbola \(4x^{2} - 25y^{2} = 100\), the equation translates to:

• \( y = \pm \frac{2}{5}x \)

These lines intersect at the hyperbola's center, which here is the origin point (0,0). Graphically, these asymptotes form an "X" pattern, guiding the curve of the hyperbola, ensuring that its branches get infinitely close to the lines without intersecting them.

Vertices of Hyperbola

Vertices are the points where the hyperbola makes its closest approach to the center along its major axis. When the hyperbola's equation is aligned with the x-axis, as in \( \frac{x^{2}}{25} - \frac{y^{2}}{4} = 1 \), the vertices are placed on the x-axis itself.

• The vertices can be calculated using the formula: \((h \pm a, k)\).
• For our hyperbola centered at (0,0), with \(a = 5\), the vertices are located at (5, 0) and (-5, 0).

These points are not just essential for graphing but also for understanding the scale and direction in which the hyperbola stretches. The distance between these vertices helps students visualize and construct the hyperbola accurately on a graph, especially when combined with its asymptotes.

Foci of Hyperbola

The foci are special points located along the major axis of the hyperbola, and they play a fundamental role in its definition and properties. The distance from the center to each focus is denoted by \(c\), which can be calculated by:

• \( c = \sqrt{a^2 + b^2} \)

For the hyperbola in question, \(4x^{2} - 25y^{2} = 100\), calculating \(c\) gives us \( \sqrt{29} \). Therefore, the coordinates of the hyperbola's foci become:

• (0 \pm \sqrt{29}, 0) \text{, which approximately maps to (5.39, 0) and (-5.39, 0)}.

Unlike the vertices, the foci are not points on the hyperbola itself but are rather pivotal in calculating other properties of a hyperbola. Their presence is integral to the mathematical definition of a hyperbola: the absolute distance of a point on the hyperbola from the foci is constant, highlighting their importance in the overall geometry.