Problem 23 - Thomas Calculus

Question

In Exercises \(17-24\) , find the eccentricity of the hyperbola. Then find and graph the hyperbola's foci and directrices. $$8 y^{2}-2 x^{2}=16$$

Step-by-Step Solution

Answer

The eccentricity is \(\sqrt{5}\). Foci: \((0, \pm \sqrt{10})\), Directrices: \(y = \pm \frac{\sqrt{10}}{5}\).

Step 1: Write the equation in standard form

The given equation is \(8y^2 - 2x^2 = 16\). Divide all terms by 16 to simplify it: \[\frac{8y^2}{16} - \frac{2x^2}{16} = 1\] Simplify the fractions: \[\frac{y^2}{2} - \frac{x^2}{8} = 1\] This is the standard form of a hyperbola.

Step 2: Determine the values of a, b, and c

In the standard form \[\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\], we identify \(a^2 = 2\) and \(b^2 = 8\). Calculate \(c^2 = a^2 + b^2\):\[c^2 = 2 + 8 = 10\] Thus, \(c = \sqrt{10}\).

Step 3: Find the eccentricity

The eccentricity \(e\) of a hyperbola is given by \(e = \frac{c}{a}\). Substitute the known values:\[e = \frac{\sqrt{10}}{\sqrt{2}} = \sqrt{\frac{10}{2}} = \sqrt{5}\].

Step 4: Determine the foci

Foci of the hyperbola are at \((0, \pm c)\) since the hyperbola is oriented vertically (\(y\)-term first). The coordinates of the foci are \((0, \pm \sqrt{10})\).

Step 5: Determine the directrices

The directrices of a hyperbola are lines given by \(y = \pm \frac{a^2}{c}\). Calculate this value:\[\frac{a^2}{c} = \frac{2}{\sqrt{10}} = \frac{2\sqrt{10}}{10} = \frac{\sqrt{10}}{5}\]. So, the directrices are \(y = \pm \frac{\sqrt{10}}{5}\).

Step 6: Sketch the hyperbola

To graph the hyperbola, mark the center at the origin \((0,0)\), the vertices at \((0, \pm \sqrt{2})\), the foci at \((0, \pm \sqrt{10})\), and draw the rectangular box using \(a\) and \(b\) values to aid in sketching the asymptotes. Sketch the hyperbola passing through the vertices and asymptotes.

Key Concepts

Hyperbola

A hyperbola is a type of conic section that results when a plane cuts through both nappes of a cone. Unlike ellipses and circles, hyperbolas have two disconnected curves called branches. These curves mirror each other and extend indefinitely. The key feature of a hyperbola is its two arms that spread in opposite directions.

When you think of a hyperbola, it's important to remember:

• It has two foci, points from which distances are measured.
• It has directrices, lines used in its geometric definition.
• The distance between the foci is greater than the distance between any point on the hyperbola and either focus.

Understanding hyperbolas at their core helps in visualizing and working with their equations.

Equation of Hyperbola

The equation of a hyperbola can present itself in several forms, but the most common is the standard form. For a hyperbola centered at the origin, the equation exhibits one of two primary forms:

• Vertical hyperbola: \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\)
• Horizontal hyperbola: \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\)

The term presenting with the positive sign dictates the orientation, whether vertical or horizontal. In our exercise, the given hyperbola is vertical because the \(y\)-term appears first and positive in the standard equation.

Identifying whether a hyperbola is vertical or horizontal is crucial for further calculations, like finding vertices, foci, and sketching the graph. This classification sets the groundwork for determining features of the hyperbola and aids in problem-solving steps.

Foci and Directrices of Hyperbola

The foci and directrices are fundamental in studying hyperbolas as they help define the geometric properties of these curves. The foci are points located along the transverse axis, which is the axis where the two branches open outwards.

In any hyperbola:

• The distance between the center and each focus is \(c\), where \(c = \sqrt{a^2 + b^2}\). For our problem, the foci are at \((0, \pm \sqrt{10})\).
• The directrices of the hyperbola provide another axis of symmetry. They are located at \(y = \pm \frac{a^2}{c}\), equating here to \(y = \pm \frac{\sqrt{10}}{5}\).

These components are not just theoretical but practical, providing locations for focusing conic sections' unique set of reflective properties.

Standard Form of a Hyperbola

Starting from the given equation, to convert the hyperbola into standard form, you divide all terms to set the equation equal to one. From our problem, we adjust and simplify into standard form, \(\frac{y^2}{2} - \frac{x^2}{8} = 1\).

Key steps include:

• Divide each term by the constant on the right side of the equation to make it equal to 1.
• Keep track of coefficients with squares under \(y\) and \(x\) terms - these inform the values of \(a\) and \(b\).

By doing so, the standard form becomes a guideline, enabling you to recognize the hyperbola's orientation and to derive its essential components, like eccentricity, foci, and directrices.