Problem 56 - A Graphical Approach to Precalculus with Limits

Question

Write an equation for each conic. Each parabola has vertex at the origin, and each ellipse or hyperbola is centered at the origin. $$\text { Focus }(2,0) ; e=\frac{6}{5}$$

Step-by-Step Solution

Answer

The equation of the hyperbola is \( \frac{9x^2}{25} - \frac{9y^2}{11} = 1 \).

Step 1: Identify the Conic Type

The given eccentricity \( e = \frac{6}{5} \) is greater than \( 1 \), which indicates that the conic section is a hyperbola.

Step 2: Determine the Orientation

Since the focus is at \((2,0)\), the transverse axis is horizontal. Therefore, the hyperbola will have the equation of the form: \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \).

Step 3: Use the Eccentricity Formula

For a hyperbola, the eccentricity \( e \) is defined as \( e = \frac{c}{a} \), where \( c \) is the distance from the center to the focus, \( c = 2 \). So, \( \frac{6}{5} = \frac{2}{a} \).

Step 4: Solve for \( a \)

From \( \frac{6}{5} = \frac{2}{a} \), solve for \( a \):\[ a = \frac{2 imes 5}{6} = \frac{10}{6} = \frac{5}{3} \].

Step 5: Calculate \( b^2 \) using Hyperbola Formula

Recall that for each hyperbola \( c^2 = a^2 + b^2 \). We have \( a = \frac{5}{3} \), so \( a^2 = \left(\frac{5}{3}\right)^2 = \frac{25}{9} \) and \( c^2 = 2^2 = 4 \).Substitute into the equation:\[ 4 = \frac{25}{9} + b^2 \].

Step 6: Solve for \( b^2 \)

Rearrange the equation to solve for \( b^2 \):\[ b^2 = 4 - \frac{25}{9} = \frac{36}{9} - \frac{25}{9} = \frac{11}{9} \].

Step 7: Write the Equation of the Hyperbola

With \( a^2 = \frac{25}{9} \) and \( b^2 = \frac{11}{9} \), the standard form of the equation is:\[ \frac{x^2}{\frac{25}{9}} - \frac{y^2}{\frac{11}{9}} = 1 \],which simplifies to:\[ \frac{9x^2}{25} - \frac{9y^2}{11} = 1 \].

Key Concepts

Hyperbola

A hyperbola is a type of conic section that is formed by the intersection of a plane with both nappes of a cone. Unlike an ellipse, a hyperbola opens outward in two opposite directions. A simple way to remember its shape is to think of it as two mirrored curves resembling open-ended U shapes.

• A hyperbola has two branches, each resembling half of a parabola but with more symmetry.
• It is defined by two focal points, and the difference in the distances from these focal points to any point on the hyperbola is constant.
• The standard form of a hyperbola centered at the origin depends on its orientation.

For a horizontal hyperbola (transverse axis along the x-axis), the standard equation is:\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]For a vertical hyperbola (transverse axis along the y-axis), the equation becomes:\[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \]Understanding this distinct structure is key to working with hyperbolas.

Eccentricity

Eccentricity, denoted as \( e \), is a parameter that defines the shape of a conic section. It is a measure of how much a conic deviates from being circular.

• For a hyperbola, the eccentricity is always greater than 1.
• This parameter helps distinguish conic sections: \( e = 0 \) for circles, \( 0 < e < 1 \) for ellipses, \( e = 1 \) for parabolas, and \( e > 1 \) for hyperbolas.

In the context of hyperbolas, eccentricity is calculated using the formula \( e = \frac{c}{a} \), where \( c \) represents the distance from the center to the focus, and \( a \) is the distance from the center to a vertex on the transverse axis.

• An eccentricity greater than one indicates a hyperbola and suggests that the curves open wide.
• The given example with \( e = \frac{6}{5} \) confirms that the conic is indeed a hyperbola as \( \frac{6}{5} > 1 \).

By understanding eccentricity, students can better predict and analyze the behavior and properties of the conic section they are working with.

Transverse Axis

The transverse axis of a hyperbola is a key feature that defines its orientation and shape. It is the axis that passes through both foci and is the line segment between the vertices.

• In a horizontal hyperbola, the transverse axis lies along the x-axis.
• In a vertical hyperbola, it lies along the y-axis.
• This axis provides a fundamental basis for determining the equation of the hyperbola.

The transverse axis helps in distinguishing the direction in which a hyperbola opens and contributes to the calculated lengths \( a \) and \( b \), crucial for forming the hyperbola's equation.For the hyperbola given in the exercise, the transverse axis is horizontal, as indicated by the focus point \((2,0)\). This orientation suggests that the equation takes the form:\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]The transverse axis is crucial because it helps in accurately plotting and understanding the true shape and direction of a hyperbola.