Algebra II worksheet with questions on conic sections and multiple-choice answers.

Problem Analysis:

The task involves solving a series of questions related to conic sections, specifically ellipses and hyperbolas. Let's address each question step by step.

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Question 1: Write the definition of a parabola.

A parabola is defined as the set of all points in a plane that are equidistant from a fixed point (called the focus) and a fixed line (called the directrix). Mathematically, if \( F \) is the focus and \( l \) is the directrix, then for any point \( P \) on the parabola, the distance from \( P \) to \( F \) is equal to the perpendicular distance from \( P \) to \( l \).

Answer: A parabola is the set of all points in a plane that are equidistant from a fixed point (focus) and a fixed line (directrix).

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Question 2: Write the definition of an ellipse.

An ellipse is defined as the set of all points in a plane such that the sum of the distances from any point on the ellipse to two fixed points (called the foci) is constant. Mathematically, if \( F_1 \) and \( F_2 \) are the foci, then for any point \( P \) on the ellipse, the sum \( PF_1 + PF_2 \) is a constant value.

Answer: An ellipse is the set of all points in a plane such that the sum of the distances from any point on the ellipse to two fixed points (foci) is constant.

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Question 3: Write the general equation for a circle with center at \((h, k)\) and radius \( r \).

The general equation of a circle with center \((h, k)\) and radius \( r \) is given by:
\[
(x - h)^2 + (y - k)^2 = r^2
\]

Answer: The general equation for a circle with center \((h, k)\) and radius \( r \) is \((x - h)^2 + (y - k)^2 = r^2\).

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Multiple Choice Questions

#### Question 4: Which of the following is the standard form of the circle \( x^2 + y^2 - 6x + 8y + 9 = 0 \)?
To find the standard form, we complete the square for both \( x \) and \( y \).

1. Start with the given equation:
\[
x^2 + y^2 - 6x + 8y + 9 = 0
\]

2. Group the \( x \)-terms and \( y \)-terms:
\[
(x^2 - 6x) + (y^2 + 8y) + 9 = 0
\]

3. Complete the square for \( x \):
\[
x^2 - 6x \quad \text{becomes} \quad (x - 3)^2 - 9
\]

4. Complete the square for \( y \):
\[
y^2 + 8y \quad \text{becomes} \quad (y + 4)^2 - 16
\]

5. Substitute back into the equation:
\[
(x - 3)^2 - 9 + (y + 4)^2 - 16 + 9 = 0
\]

6. Simplify:
\[
(x - 3)^2 + (y + 4)^2 - 16 = 0
\]
\[
(x - 3)^2 + (y + 4)^2 = 16
\]

The standard form of the circle is:
\[
(x - 3)^2 + (y + 4)^2 = 16
\]

Answer: D. \((x - 3)^2 + (y + 4)^2 = 16\)

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#### Question 5: What do we call the solution of the equation?
The equation given is:
\[
\begin{cases}
x^2 + y^2 = 4 \\
x^2 + y^2 = 9
\end{cases}
\]

These are equations of two circles:
1. \( x^2 + y^2 = 4 \) (radius 2, centered at \((0, 0)\))
2. \( x^2 + y^2 = 9 \) (radius 3, centered at \((0, 0)\))

Since both circles share the same center but have different radii, they do not intersect. Therefore, there are no common solutions.

Answer: B. No solution

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#### Question 6: Identify the conic section represented by the equation \( x^2 + 4y^2 + 6x - 8y + 9 = 0 \).
To identify the conic section, we rewrite the equation in standard form by completing the square.

1. Start with the given equation:
\[
x^2 + 4y^2 + 6x - 8y + 9 = 0
\]

2. Group the \( x \)-terms and \( y \)-terms:
\[
(x^2 + 6x) + 4(y^2 - 2y) + 9 = 0
\]

3. Complete the square for \( x \):
\[
x^2 + 6x \quad \text{becomes} \quad (x + 3)^2 - 9
\]

4. Complete the square for \( y \):
\[
y^2 - 2y \quad \text{becomes} \quad (y - 1)^2 - 1
\]

5. Substitute back into the equation:
\[
(x + 3)^2 - 9 + 4((y - 1)^2 - 1) + 9 = 0
\]

6. Simplify:
\[
(x + 3)^2 - 9 + 4(y - 1)^2 - 4 + 9 = 0
\]
\[
(x + 3)^2 + 4(y - 1)^2 - 4 = 0
\]
\[
(x + 3)^2 + 4(y - 1)^2 = 4
\]

7. Divide through by 4 to get the standard form:
\[
\frac{(x + 3)^2}{4} + \frac{(y - 1)^2}{1} = 1
\]

This is the standard form of an ellipse:
\[
\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1
\]

Answer: C. Ellipse

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#### Question 7: Identify the conic section represented by the equation \( x^2 - 4y^2 + 6x - 8y + 9 = 0 \).
To identify the conic section, we rewrite the equation in standard form by completing the square.

1. Start with the given equation:
\[
x^2 - 4y^2 + 6x - 8y + 9 = 0
\]

2. Group the \( x \)-terms and \( y \)-terms:
\[
(x^2 + 6x) - 4(y^2 + 2y) + 9 = 0
\]

3. Complete the square for \( x \):
\[
x^2 + 6x \quad \text{becomes} \quad (x + 3)^2 - 9
\]

4. Complete the square for \( y \):
\[
y^2 + 2y \quad \text{becomes} \quad (y + 1)^2 - 1
\]

5. Substitute back into the equation:
\[
(x + 3)^2 - 9 - 4((y + 1)^2 - 1) + 9 = 0
\]

6. Simplify:
\[
(x + 3)^2 - 9 - 4(y + 1)^2 + 4 + 9 = 0
\]
\[
(x + 3)^2 - 4(y + 1)^2 + 4 = 0
\]
\[
(x + 3)^2 - 4(y + 1)^2 = -4
\]

7. Divide through by \(-4\) to get the standard form:
\[
\frac{(x + 3)^2}{-4} - \frac{(y + 1)^2}{1} = 1
\]
\[
\frac{(y + 1)^2}{1} - \frac{(x + 3)^2}{4} = 1
\]

This is the standard form of a hyperbola:
\[
\frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1
\]

Answer: D. Hyperbola

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Final Answers:

1. A parabola is the set of all points in a plane that are equidistant from a fixed point (focus) and a fixed line (directrix).
2. An ellipse is the set of all points in a plane such that the sum of the distances from any point on the ellipse to two fixed points (foci) is constant.
3. The general equation for a circle with center \((h, k)\) and radius \( r \) is \((x - h)^2 + (y - k)^2 = r^2\).
4. D. \((x - 3)^2 + (y + 4)^2 = 16\)
5. B. No solution
6. C. Ellipse
7. D. Hyperbola

Boxed Final Answer:
\[
\boxed{D, B, C, D}
\]