Problem 6 - Precalculus

Question

Determine the foci, the eccentricity, and the directrices for each ellipse and hyperbola. (a) \(x^{2}+4 y^{2}=1\) (b) \(x^{2}-4 y^{2}=1\)

Step-by-Step Solution

Answer

Foci of (a): \( \pm \frac{\sqrt{3}}{2}, 0 \); Eccentricity: \( \frac{\sqrt{3}}{2} \); Directrices: \( x = \pm \frac{2}{\sqrt{3}} \). Foci of (b): \( \pm \frac{\sqrt{5}}{2}, 0 \); Eccentricity: \( \frac{\sqrt{5}}{2} \); Directrices: \( x = \pm \frac{2}{\sqrt{5}} \).

Step 1: Identify the Type of Conic Section for (a)

The equation given for (a) is \( x^2 + 4y^2 = 1 \). Both terms are positive and one coefficient is different from the other, indicating that this is an ellipse.

Step 2: Standard Equation for Ellipse

For an ellipse in the form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), \( a \) and \( b \) can be found by rewriting the equation: \( \frac{x^2}{1} + \frac{y^2}{\frac{1}{4}} = 1 \), giving \( a^2 = 1 \) and \( b^2 = \frac{1}{4} \). Thus, \( a = 1 \) and \( b = \frac{1}{2} \).

Step 3: Calculate Foci for Ellipse

The foci of an ellipse are found using \( c^2 = a^2 - b^2 \). Here, \( c^2 = 1 - \frac{1}{4} = \frac{3}{4} \), thus \( c = \frac{\sqrt{3}}{2} \). The foci are \( (\pm c, 0) \) = \( \left( \pm \frac{\sqrt{3}}{2}, 0 \right) \).

Step 4: Eccentricity of Ellipse

The eccentricity \( e \) is given by \( e = \frac{c}{a} \). Here, \( e = \frac{\frac{\sqrt{3}}{2}}{1} = \frac{\sqrt{3}}{2} \).

Step 5: Directrices of Ellipse

For an ellipse, the directrix \( x = \pm \frac{a^2}{c} \). Here, \( x = \pm \frac{1}{\frac{\sqrt{3}}{2}} = \pm \frac{2}{\sqrt{3}} \).

Step 6: Identify the Type of Conic Section for (b)

The equation given for (b) is \( x^2 - 4y^2 = 1 \). One term is positive and the other negative, indicating that this is a hyperbola.

Step 7: Standard Equation for Hyperbola

For a hyperbola in the form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), transform the equation to \( \frac{x^2}{1} - \frac{y^2}{\frac{1}{4}} = 1 \), where \( a^2 = 1 \) and \( b^2 = \frac{1}{4} \). Thus, \( a = 1 \) and \( b = \frac{1}{2} \).

Step 8: Calculate Foci for Hyperbola

The foci for a hyperbola are found using \( c^2 = a^2 + b^2 \). Here, \( c^2 = 1 + \frac{1}{4} = \frac{5}{4} \), thus \( c = \frac{\sqrt{5}}{2} \). The foci are \( (\pm c, 0) \) = \( \left( \pm \frac{\sqrt{5}}{2}, 0 \right) \).

Step 9: Eccentricity of Hyperbola

The eccentricity \( e \) is given by \( e = \frac{c}{a} \). Here, \( e = \frac{\frac{\sqrt{5}}{2}}{1} = \frac{\sqrt{5}}{2} \).

Step 10: Directrices of Hyperbola

For a hyperbola, the directrix is given by \( x = \pm \frac{a^2}{e} = \pm \frac{1}{\frac{\sqrt{5}}{2}} = \pm \frac{2}{\sqrt{5}} \).

Key Concepts

Foci of Conic Sections

The foci of conic sections, such as ellipses and hyperbolas, are essential points that help define the shape and orientation of these curves. For both ellipses and hyperbolas, the foci are located along the major axis of the conic section. For ellipses, the foci are closer together compared to hyperbolas. The formula to determine the distance of the foci from the center of an ellipse is given by: \[ c^2 = a^2 - b^2 \] where \( c \) is the distance from the center to each focus, \( a \) is the semi-major axis, and \( b \) is the semi-minor axis. In an ellipse, the foci are inside the ellipse, along the major axis, at \( (\pm c, 0) \) if the major axis is horizontal or \( (0, \pm c) \) if it is vertical. For hyperbolas, the foci are outside the vertices. The relationship is expressed as: \[ c^2 = a^2 + b^2 \] Here, \( c \) measures the distance from the center to each focus. For a hyperbola, the foci also lie along the transverse axis, at \( (\pm c, 0) \) or \( (0, \pm c) \), depending on the orientation of the transverse axis.

• The foci help in calculating the direction and elongation of ellipses and hyperbolas.
• They are crucial in understanding the interaction of light and signals with these curves, such as in satellite dishes and optical lenses.

Eccentricity Formula

Eccentricity is a measure of how much a conic section deviates from being circular. It's a crucial concept that defines whether a conic section is an ellipse, parabola, or hyperbola.The eccentricity is denoted by \( e \). For ellipses, the eccentricity formula is: \[ e = \frac{c}{a} \] where \( c \) is the distance from the center to a focus, and \( a \) is the length of the semi-major axis. For ellipses, \( 0 < e < 1 \), meaning they are more circular when \( e \) is closer to 0.For hyperbolas, the formula remains the same, \( e = \frac{c}{a} \), but in this case, \( e > 1 \). This indicates the hyperbola's branches are more spread out, making it appear less circular.

• Eccentricity tells us the shape and light reflection properties of the curve.
• An eccentricity of exactly 1 would define a parabola, another important conic shape.

Knowing the eccentricity helps in various fields such as astronomy, where it characterizes the orbit of planets and comets.

Directrix of Conic Sections

The directrix of conic sections serves as a reference line used to define the curve's shape in relation to its focus.In an ellipse, although the directrix is not frequently mentioned, it can be calculated and helps understand the geometric construction of the ellipse. For an ellipse, the directrix is perpendicular to the major axis and at a distance from the center given by: \[ x = \pm \frac{a^2}{c} \] where \( a \) is the semi-major axis length and \( c \) is the distance from the center to a focus.In hyperbolas, the directrices are much more critical since they help describe the open shape of the hyperbola. The directrix for hyperbolas is given by: \[ x = \pm \frac{a^2}{e} \] where \( e \) is the eccentricity. Here, the directrix provides a boundary condition that keeps the hyperbola's open arms extending towards infinity.

• Directrices help in geometric designs and constructions.
• They are used in practical applications like engineering designs involving reflective properties of hyperbolas.

Understanding the directrix's role can enrich a student's comprehension of conic sections and their applications.