Question

For the following exercises, determine the equation of the hyperbola using the information given. \((0,10),(0,-10)\) and eccentricity of \(2.5\)

Short answer

\(\frac{y^2}{100} - \frac{x^2}{525} = 1\).

Step-by-step solution

Determine the Properties

The given points \((0, 10)\) and \((0, -10)\) are the vertices of the hyperbola. Since the \(y\)-coordinates differ, the hyperbola's axis is vertical.

Calculate the Distance Between Vertices

The distance between the vertices is \(20\) because it's the difference in \(y\)-coordinates: \(10 - (-10) = 20\). The length of the transverse axis is thus \(2a = 20\), so \(a = 10\).

Use Eccentricity to Find \(c\)

The eccentricity \(e\) is given as \(2.5\), and we know \(e = \frac{c}{a}\). Using \(a = 10\), substitute to find \(c\): \[2.5 = \frac{c}{10}\] \[c = 25\].

Calculate \(b\) Using \(c^2 = a^2 + b^2\)

Now that we have \(c = 25\) and \(a = 10\), find \(b^2\) using the relation \(c^2 = a^2 + b^2\):\[c^2 = 25^2 = 625\]\[a^2 = 10^2 = 100\]\[625 = 100 + b^2\]\[b^2 = 525\].

Write the Standard Equation of the Hyperbola

For a vertical hyperbola centered at the origin with \(a = 10\) and \(b^2 = 525\), the standard form will be:\[\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\]Substitute \(a^2 = 100\) and \(b^2 = 525\):\[\frac{y^2}{100} - \frac{x^2}{525} = 1\].

Key concepts

Vertices of Hyperbola

Hyperbolas are defined by specific points, known as vertices, which are the turning points of the curve. In our problem, these vertices are located at the points

• \((0, 10)\)
• \((0, -10)\)

The vertices show that the hyperbola opens along the y-axis since they have the same x-coordinate.
This means that instead of a left-to-right opening, it opens vertically. The distance between them is essential to determine further properties, like the transverse axis, which measures 20 units in total.
By calculating the distance between the vertices, we find the parameter \(a\) of the hyperbola is 10, using the formula:

• \(2a = 20\)
• \(a = \frac{20}{2} = 10\)

Knowing the vertices allows establishing the orientation and basic dimensions of the hyperbola.

Eccentricity of Hyperbola

Eccentricity, denoted by the letter \(e\), is a crucial concept in understanding conic sections like hyperbolas. It describes how much the conic section deviates from being a circle. For hyperbolas, eccentricity is always greater than 1.
In our exercise, the eccentricity is given as 2.5. We use this value to find other properties of the hyperbola, such as the focal distance, \(c\). The relationship between the eccentricity, \(a\), and \(c\) is given by:

• \(e = \frac{c}{a}\)

Given that \(a = 10\), we can calculate \(c\) as follows:

• \(2.5 = \frac{c}{10}\)
• \(c = 2.5 \times 10 = 25\)

This means the foci of the hyperbola, points indicating where it is most elongated, are located 25 units from the center, both above and below.

Standard Form of Hyperbola

The standard form of a hyperbola helps describe its geometric properties using an algebraic equation. For hyperbolas centered at the origin, you can recognize the form based on the axis orientation:

• If the hyperbola opens vertically, the standard form is: \[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \]
• If the hyperbola opens horizontally, it is: \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]

In this problem, the hyperbola opens along the vertical axis, so we use the first equation. With parameters identified as \(a^2 = 100\) (since \(a = 10\)) and \(b^2 = 525\), we construct the hyperbola's equation:

• \[ \frac{y^2}{100} - \frac{x^2}{525} = 1 \]

This gives a complete description of our vertical hyperbola centered at the origin. By understanding this equation, you can easily visualize and characterize the hyperbola's structure and orientation based on key dimensions.