Eccentricity of Conic Sections worksheet for algebra practice.

To solve the problem of identifying the eccentricity of each conic section, we need to analyze each equation and determine whether it represents a circle, ellipse, parabola, or hyperbola. The eccentricity \( e \) is defined differently for each type of conic section:

1. Circle: \( e = 0 \)
2. Ellipse: \( 0 < e < 1 \)
3. Parabola: \( e = 1 \)
4. Hyperbola: \( e > 1 \)

Step-by-Step Solution:

#### 1. \( x^2 + y^2 = 1 \)
This is the equation of a circle with radius 1 centered at the origin. For a circle, the eccentricity is:
\[ e = 0 \]

#### 2. \( x^2 + \frac{y^2}{4} = 1 \)
This is the equation of an ellipse. The standard form of an ellipse is:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
Here, \( a^2 = 1 \) and \( b^2 = 4 \). Since \( b^2 > a^2 \), the major axis is vertical. The eccentricity \( e \) of an ellipse is given by:
\[ e = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]

#### 3. \( \frac{x^2}{9} + y^2 = 1 \)
This is also the equation of an ellipse. Here, \( a^2 = 9 \) and \( b^2 = 1 \). Since \( a^2 > b^2 \), the major axis is horizontal. The eccentricity \( e \) is:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{1}{9}} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3} \]

#### 4. \( x^2 + y^2 = 8 \)
This is the equation of a circle with radius \( \sqrt{8} \). The eccentricity is:
\[ e = 0 \]

#### 5. \( \frac{x^2}{4} + \frac{y^2}{9} = 1 \)
This is the equation of an ellipse. Here, \( a^2 = 9 \) and \( b^2 = 4 \). Since \( a^2 > b^2 \), the major axis is vertical. The eccentricity \( e \) is:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3} \]

#### 6. \( (x+2)^2 + (y-3)^2 = 4 \)
This is the equation of a circle with center \((-2, 3)\) and radius 2. The eccentricity is:
\[ e = 0 \]

#### 7. \( (x+y)(x-y) = 8 \)
This simplifies to:
\[ x^2 - y^2 = 8 \]
This is the equation of a hyperbola. The standard form of a hyperbola is:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
Here, \( a^2 = 8 \) and \( b^2 = 8 \). The eccentricity \( e \) of a hyperbola is given by:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{8}{8}} = \sqrt{2} \]

#### 8. \( 2x^2 + 3y^2 = 6 \)
This can be rewritten as:
\[ \frac{x^2}{3} + \frac{y^2}{2} = 1 \]
This is the equation of an ellipse. Here, \( a^2 = 3 \) and \( b^2 = 2 \). Since \( a^2 > b^2 \), the major axis is horizontal. The eccentricity \( e \) is:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{2}{3}} = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} \]

#### 9. \( x^2 - 2x + y^2 + 8 = 0 \)
Complete the square for \( x \):
\[ x^2 - 2x + 1 + y^2 = -8 + 1 \]
\[ (x-1)^2 + y^2 = -7 \]
This is not a valid equation for a conic section because the right-hand side is negative. Therefore, there is no conic section here.

#### 10. \( 2x^2 + 3y^2 + 8xy + 8 = 0 \)
This is a general second-degree equation. To identify the conic section, we need to check the discriminant \( B^2 - 4AC \), where \( A = 2 \), \( B = 8 \), and \( C = 3 \):
\[ B^2 - 4AC = 8^2 - 4 \cdot 2 \cdot 3 = 64 - 24 = 40 \]
Since \( B^2 - 4AC > 0 \), this represents a hyperbola. However, without further simplification, we cannot directly find the eccentricity.

Final Answer:

\[
\boxed{
\begin{array}{ll}
1. & 0 \\
2. & \frac{\sqrt{3}}{2} \\
3. & \frac{2\sqrt{2}}{3} \\
4. & 0 \\
5. & \frac{\sqrt{5}}{3} \\
6. & 0 \\
7. & \sqrt{2} \\
8. & \frac{\sqrt{3}}{3} \\
9. & \text{No conic section} \\
10. & \text{Hyperbola (eccentricity not directly computable without further simplification)}
\end{array}
}
\]