The problem is to identify the type of conic section represented by each of the given equations. The equations are: 3. $\frac{x^2}{9} + \frac{y^2}{4} = 1$ 4. $-\frac{x^2}{9} + \frac{y^2}{4} = -1$ 5. $-\frac{x^2}{9} + \frac{y}{4} = 0$ 6. $9x^2 + 4y^2 = 9$ 7. $x^2 - 4y^2 = 4$

1. Problem Description

The problem is to identify the type of conic section represented by each of the given equations. The equations are:

3. $\frac{x^2}{9} + \frac{y^2}{4} = 1$

4. $-\frac{x^2}{9} + \frac{y^2}{4} = -1$

5. $-\frac{x^2}{9} + \frac{y}{4} = 0$

6. $9x^2 + 4y^2 = 9$

7. $x^2 - 4y^2 = 4$

2. Solution Steps

3. $\frac{x^2}{9} + \frac{y^2}{4} = 1$

This equation is of the form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, which represents an ellipse. Since $a^2 = 9$ and $b^2 = 4$, we have $a = 3$ and $b = 2$. Since $a > b$, this is a horizontal ellipse.

4. $-\frac{x^2}{9} + \frac{y^2}{4} = -1$

Multiplying the equation by -1 gives $\frac{x^2}{9} - \frac{y^2}{4} = 1$. This is of the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, which represents a horizontal hyperbola.

5. $-\frac{x^2}{9} + \frac{y}{4} = 0$

This can be rewritten as $\frac{y}{4} = \frac{x^2}{9}$, or $y = \frac{4}{9}x^2$. This is a parabola that opens upwards.

6. $9x^2 + 4y^2 = 9$

Divide by 9 to get $x^2 + \frac{4y^2}{9} = 1$, which can be rewritten as $\frac{x^2}{1} + \frac{y^2}{\frac{9}{4}} = 1$, or $\frac{x^2}{1^2} + \frac{y^2}{(\frac{3}{2})^2} = 1$. This represents an ellipse. Since $\frac{3}{2} > 1$, this is a vertical ellipse.

7. $x^2 - 4y^2 = 4$

Divide by 4 to get $\frac{x^2}{4} - \frac{4y^2}{4} = 1$, which simplifies to $\frac{x^2}{4} - \frac{y^2}{1} = 1$.

This is of the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, which represents a hyperbola. Since the $x^2$ term is positive, this is a horizontal hyperbola.

3. Final Answer

3. Horizontal ellipse

4. Horizontal hyperbola

5. Parabola

6. Vertical ellipse

7. Horizontal hyperbola