pre cal 1st

Conic Section: A curve obtained as the intersection of the surface of a cone with a plane.
Nappes: The two circular structures forming the surface of a cone used to define conic sections.
Circle: All points that are equidistant from a central point. The distance from the center point $A$ to any point on the circle is defined as the radius ( $r$ ).
Ellipse: All points found by keeping the sum of the distances from two specific points constant. It is visually described as an oval.
Hyperbola: All points found by keeping the difference of the distances from two specific points constant surface.

Definition: A curve where every point is equidistant from a fixed point called the focus and a fixed line called the directrix.
Focus: A fixed point denoted as point $A$ .
Vertex: The peak or highest/lowest point of the quadratic function.
Directrix: A fixed line used in the geometric construction of the parabola.
Mathematical Context: Often associated with a quadratic function.

Formula: The distance $C$ between two points in a rectangular coordinate system is calculated using: $C = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$
Example Application: To find the distance between the points $(6, 4)$ and $(2, 1)$ :
- Identify coordinates: $(x_1, y_1) = (6, 4)$ and $(x_2, y_2) = (2, 1)$ .
- Substitute into formula: $C = \sqrt{(2 - 6)^2 + (1 - 4)^2}$
- Simplify: $C = \sqrt{(-4)^2 + (-3)^2}$
- $C = \sqrt{16 + 9}$
- $C = \sqrt{25}$
- $C = 5$

Definition: The midpoint is the center point of a line segment whose endpoints are defined by two specific points ( $x_1, y_1$ ) and ( $x_2, y_2$ ).
Formula: The midpoint coordinate is calculated as: $(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})$
Example Application: Use the Midpoint Formula to find the midpoint of a line segment with endpoints $(-5, -4)$ and $(7, 2)$ .
- $x$ -coordinate calculation: $\frac{-5 + 7}{2} = \frac{2}{2} = 1$
- $y$ -coordinate calculation: $\frac{-4 + 2}{2} = \frac{-2}{2} = -1$
- Final Midpoint: $(1, -1)$

Definition: A circle radius ( $r$ ) is the distance from the center $(h, k)$ to any point on the circle $(x, y)$ .
Standard Form Equation: For a circle with center $(h, k)$ , the equation is: $(x - h)^2 + (y - k)^2 = r^2$
Example 1: Circle with radius $3$ and center $(0, 0)$ .
- Equation: $(x - 0)^2 + (y - 0)^2 = 3^2$
- Result: $x^2 + y^2 = 9$
Example 2: Circle with radius $2$ and center $(-1, 3)$ .
- Equation: $(x - (-1))^2 + (y - 3)^2 = 2^2$
- Result: $(x + 1)^2 + (y - 3)^2 = 4$
Example 3: Find the standard form of the equation of a circle with center $(2, 4)$ that also contains the point $(-2, 1)$ .

General Formula: The expanded equation of a circle is expressed as: $x^2 + y^2 - ax - by + c = 0$
Finding Center and Radius from Given Equations:
- Case 1: $4x^2 + 4y^2 = 64$
  - Divide by $4$ : $x^2 + y^2 = 16$
  - Center: $(0, 0)$
  - Radius ( $r$ ): $r^2 = 16 \rightarrow r = 4$
- Case 2: $3x^2 + 3y^2 = 27$
  - Divide by $3$ : $x^2 + y^2 = 9$
  - Center: $(0, 0)$
  - Radius ( $r$ ): $r^2 = 9 \rightarrow r = 3$
- Case 3: Finding center and radius for $x^2 + y^2 - 4x - 6y + 4 = 0$
  - Rearrange: $(x^2 - 4x) + (y^2 - 6y) = -4$
  - Complete the square for $x$ : $\frac{-4}{2} = -2 \rightarrow (-2)^2 = 4$
  - Complete the square for $y$ : $\frac{-6}{2} = -3 \rightarrow (-3)^2 = 9$
  - Resulting Equation: $(x^2 - 4x + 4) + (y^2 - 6y + 9) = -4 + 4 + 9$
  - Standard Form: $(x - 2)^2 + (y - 3)^2 = 9$
  - Center: $(2, 3)$ , Radius: $3$

Fixed Points and Lines:
- Focus: Fixed point.
- Directrix: Fixed line.
- Axis of Symmetry: Line passing through the vertex and focus.
- Latus Rectum: Line segment through the focus, perpendicular to the axis of symmetry, with endpoints on the parabola.
Parabolas with Vertex at the Origin (0,0):
- Horizontal Axis of Symmetry ( $x$ -axis):
  - Equation: $y^2 = 4px$ (where $p \neq 0$ )
  - Focus: $(p, 0)$
  - Directrix: $x = -p$
  - Endpoints of Latus Rectum: $(p, \pm 2p)$
  - Direction: Opens right if p > 0, opens left if p < 0.
- Vertical Axis of Symmetry ( $y$ -axis):
  - Equation: $x^2 = 4py$ (where $p \neq 0$ )
  - Focus: $(0, p)$
  - Directrix: $y = -p$
  - Endpoints of Latus Rectum: $(\pm 2p, p)$
  - Direction: Opens up if p > 0, opens down if p < 0.

Problem: Graph $y^2 = 24x$ . Identify focus, directrix, and latus rectum endpoints.
1. Compare to standard form $y^2 = 4px$ .
2. Calculate $p$ : $4p = 24 \rightarrow p = 6$ .
3. Because $p$ is positive ( $6$ ), it opens to the right.
4. Vertex: $(0,0)$ .
5. Focus: $(p, 0) = (6, 0)$ .
6. Directrix: $x = -p \rightarrow x = -6$ .
7. Endpoints of Latus Rectum: $(p, \pm 2p) = (6, \pm 12)$ . Points: $(6, 12)$ and $(6, -12)$ .
Problem: Graph $x^2 = -6y$ .
1. Compare to standard form $x^2 = 4py$ .
2. Calculate $p$ : $4p = -6 \rightarrow p = -\frac{6}{4} = -\frac{3}{2}$ .
3. Focus: $(0, p) = (0, -\frac{3}{2})$ .
4. Directrix: $y = -p \rightarrow y = \frac{3}{2}$ .
5. Endpoints of Latus Rectum: $(\pm 2p, p) = (\pm 2(-\frac{3}{2}), -\frac{3}{2}) = (\pm 3, -\frac{3}{2})$ . Points: $(3, -\frac{3}{2})$ and $(-3, -\frac{3}{2})$ .
6. Directional rules: If $x$ is positive (right), negative (left); if $y$ is positive (up), negative (down).

Summary Table for Vertex (h, k):
- Equation Horizontal: $(y-k)^2 = 4p(x-h)$
  - Axis of Symmetry: $y = k$
  - Focus: $(h+p, k)$
  - Directrix: $x = h - p$
  - Endpoints of Latus Rectum: $(h+p, k \pm 2p)$
- Equation Vertical: $(x-h)^2 = 4p(y-k)$
  - Axis of Symmetry: $x = h$
  - Focus: $(h, k+p)$
  - Directrix: $y = k - p$
  - Endpoints of Latus Rectum: $(h \pm 2p, k+p)$
Example: $(y-1)^2 = -16(x+3)$
1. Vertex $(h, k)$ : $(-3, 1)$ .
2. Calculate $p$ : $4p = -16 \rightarrow p = -4$ .
3. Direction: Opens left (since $p$ is negative and axis is horizontal).
4. Axis of Symmetry: $y = 1$ .
5. Focus: $(h+p, k) = (-3 + (-4), 1) = (-7, 1)$ .
6. Directrix: $x = h - p = -3 - (-4) \rightarrow x = 1$ .
7. Endpoints of Latus Rectum: $(h+p, k \pm 2p) = (-7, 1 \pm 8)$ , resulting in points $(-7, 9)$ and $(-7, -7)$ .