Conic Sections: Circles, Ellipses, Hyperbolas, and Parabolas - Comprehensive Study Guide

Conic Sections Overview

• Definition: A conic section is a curve obtained by the intersection of a plane with a cone. The primary types are circles, ellipses, hyperbolas, and parabolas.
• Study Guide Goals:
  • Identify and graph each conic section based on its standard equation.
  • Convert general equations into standard form using "completing the square."
  • Extract critical features: centers, vertices, co-vertices, foci, axes, and radii.
  • Apply core formulas for focal distances and semi-axes.
  • Utilize a decision tree for rapid conic type identification from general forms.

Graphing Circles

• Standard Form of a Circle: The equation is given by $(x - h)^2 + (y - k)^2 = r^2$.
  • Center: $(h, k)$.
  • Radius: $r$.
• Example 1: Circle Centered at Origin:
  • Equation: $x^2 + y^2 = 9$.
  • Identification: Here, $h = 0$, $k = 0$, and $r^2 = 9$.
  • Features: Center is $(0, 0)$, Radius $r = 3$.
  • Steps: Start at $(0, 0)$, move $3$ units right, left, up, and down. Connect the four points.
• Example 2: Shifted Circle:
  • Equation: $(x - 3)^2 + (y + 4)^2 = 16$.
  • Identification: The signs in the equation are negative relative to the form, so $(x - 3)$ implies $h = 3$ and $(y + 4)$ implies $k = -4$.
  • Features: Center is $(3, -4)$. Radius $r = \sqrt{16} = 4$.
  • Steps: Plot center at $(3, -4)$, move $4$ units in cardinal directions, then connect.
• Example 3: Negative Coordinates:
  • Equation: $(x + 2)^2 + (y - 3)^2 = 4$.
  • Identification: $h = -2$, $k = 3$.
  • Features: Center is $(-2, 3)$. Radius $r = 2$.

Graphing Ellipses

• Standard Forms of an Ellipse:

  • Horizontal: $\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$ (where $a > b$).
  • Vertical: $\frac{(x - h)^2}{b^2} + \frac{(y - k)^2}{a^2} = 1$ (where $a > b$).
  • Rule: In an ellipse, $a^2$ is always the larger denominator. If the larger denominator is under $x$, the ellipse is horizontal; if it is under $y$, the ellipse is vertical.

• Key Measurements:

  • Center: $(h, k)$.
  • Semi-major axis ($a$): Distance from center to vertices (the longer axis).
  • Semi-minor axis ($b$): Distance from center to co-vertices (the shorter axis).
  • Focal distance ($c$): Calculated using $c^2 = a^2 - b^2$. Foci lie on the major axis inside the ellipse.
  • Major Axis Length: $2a$.
  • Minor Axis Length: $2b$.

• Example 1: Vertical Ellipse (Origin):

  • Equation: $\frac{x^2}{25} + \frac{y^2}{49} = 1$.
  • Identification: Larger denominator is under $y$, so $a^2 = 49$ and $b^2 = 25$. Center is $(0, 0)$.
  • Features: $a = 7$, $b = 5$. Major axis length $= 14$, Minor axis length $= 10$.
  • Foci: $c^2 = 49 - 25 = 24$, so $c = \sqrt{24} = 2\sqrt{6} \approx 4.89$. Foci are at $(0, \pm 4.89)$.

• Example 2: Horizontal Ellipse (Shifted):

  • Equation: $\frac{(x - 2)^2}{16} + \frac{(y + 3)^2}{9} = 1$.
  • Identification: Larger denominator is under $x$, so $a^2 = 16$ and $b^2 = 9$.
  • Features: Center is $(2, -3)$. $a = 4$, $b = 3$.
  • Vertices: $(h \pm a, k) = (2 \pm 4, -3) \rightarrow (6, -3)$ and $(-2, -3)$.
  • Co-vertices: $(h, k \pm b) = (2, -3 \pm 3) \rightarrow (2, 0)$ and $(2, -6)$.
  • Foci: $c = \sqrt{16 - 9} = \sqrt{7}$. Foci are at $(2 \pm \sqrt{7}, -3)$.

• Example 3: Vertical Ellipse (Shifted):

  • Equation: $\frac{(x + 1)^2}{9} + \frac{(y - 2)^2}{25} = 1$.
  • Identification: $a^2 = 25$ (under $y$), $b^2 = 9$, Center is $(-1, 2)$.
  • Features: $a = 5$, $b = 3$. $c = \sqrt{25 - 9} = 4$.
  • Vertices: $(-1, 2 \pm 5) \rightarrow (-1, 6)$ and $(-1, -2)$.
  • Foci: $(-1, 2 \pm 4) \rightarrow (-1, 6)$ and $(-1, -2)$. (Wait, the focus shares coordinates with vertices in this calculation if $c=a$, but here $c=4$ whereas $a=5$). Foci: $(-1, 6)$ and $(-1, -2)$.
  • Finding Intercepts: To find x-intercepts, set $y = 0$ and solve for $x$. To find y-intercepts, set $x=0$ and solve for $y$.

Graphing Hyperbolas

• Standard Forms of a Hyperbola:

  • Horizontal: $\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$ (opens left and right).
  • Vertical: $\frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1$ (opens up and down).
  • Critical Distinction: Unlike ellipses, $a^2$ is the denominator of the positive term, not necessarily the larger number.

• Key Measurements:

  • Center: $(h, k)$.
  • Transverse Axis: The axis containing the vertices.
  • Vertices: Distanced $a$ from center along the transverse axis.
  • Focal distance ($c$): Calculated using $c^2 = a^2 + b^2$. Foci lie outside the vertices.
  • Asymptote Box: Created using distance $a$ along the transverse axis and distance $b$ along the conjugate axis.
  • Asymptote Equations:
    • Horizontal: $y - k = \pm\frac{b}{a}(x - h)$.
    • Vertical: $y - k = \pm\frac{a}{b}(x - h)$.

• Example 1: Horizontal Hyperbola (Origin):

  • Equation: $\frac{x^2}{9} - \frac{y^2}{4} = 1$.
  • Features: Center $(0, 0)$. $a^2 = 9 \rightarrow a = 3$. $b^2 = 4 \rightarrow b = 2$. opens left/right.
  • Foci: $c^2 = 9 + 4 = 13 \rightarrow c = \sqrt{13}$. Foci at $(\pm \sqrt{13}, 0)$.
  • Asymptotes: Slope is rise over run. From center, up $2$ ($b$) and right $3$ ($a$). Equation $y = \pm \frac{2}{3}x$.

• Example 2: Horizontal Hyperbola (Shifted):

  • Equation: $\frac{(x - 4)^2}{25} - \frac{(y - 2)^2}{16} = 1$.
  • Features: Center $(4, 2)$. $a = 5$, $b = 4$. $c = \sqrt{25 + 16} = \sqrt{41}$.
  • Vertices: $(4 \pm 5, 2) \rightarrow (9, 2)$ and $(-1, 2)$.
  • Asymptotes: $y - 2 = \pm\frac{4}{5}(x - 4)$.

• Example 3: Vertical Hyperbola (Shifted):

  • Equation: $\frac{(y - 2)^2}{4} - \frac{(x + 3)^2}{9} = 1$.
  • Features: Center $(-3, 2)$. $a^2$ is under $y$, so $a = 2$ and $b = 3$.
  • Vertices: $(-3, 2 \pm 2) \rightarrow (-3, 4)$ and $(-3, 0)$.
  • Foci: $c = \sqrt{4 + 9} = \sqrt{13}$. Foci at $(-3, 2 \pm \sqrt{13})$.
  • Asymptotes: $y - 2 = \pm\frac{2}{3}(x + 3)$.

Graphing Parabolas

• Definition: The set of all points equidistant from a fixed point (focus) and a fixed line (directrix).

• Standard Forms:

  • Horizontal: $(y - k)^2 = 4p(x - h)$.
    • Opens right if $p > 0$. Opens left if $p < 0$.
  • Vertical: $(x - h)^2 = 4p(y - k)$.
    • Opens up if $p > 0$. Opens down if $p < 0$.

• Key Measurements:

  • Vertex: $(h, k)$.
  • Distance ($p$): Distance from vertex to focus and from vertex to directrix.
  • Focus: Always located inside the curve.
  • Directrix: Always located outside the curve.

• Example 1: Horizontal Parabola (Origin):

  • Equation: $y^2 = 8x$.
  • Identification: Horizontal ($y$ is squared). $4p = 8 \rightarrow p = 2$.
  • Features: Vertex $(0, 0)$. Focus $(2, 0)$. Directrix $x = -2$.
  • Plotting points: Solving $x = \frac{y^2}{8}$. If $y = \pm 4$, $x = \frac{16}{8} = 2$. Points: $(2, 4), (2, -4)$.

• Example 2: Shifted Horizontal Parabola:

  • Equation: $(y - 2)^2 = 4(x - 3)$.
  • Features: Vertex $(3, 2)$. $4p = 4 \rightarrow p = 1$. Opens right.
  • Focus: $(3 + 1, 2) = (4, 2)$. Directrix $x = 3 - 1 = 2$.
  • Plotting Points: $x = 3 + \frac{(y - 2)^2}{4}$. If $y = 4$, $x = 3 + 1 = 4$. If $y = 0$, $x = 4$.

• Example 3: Vertical Parabola Opening Down:

  • Equation: $(x + 1)^2 = -2(y - 3)$.
  • Features: Vertex $(-1, 3)$. $4p = -2 \rightarrow p = -0.5$. Opens down.
  • Focus: $(-1, 3 - 0.5) = (-1, 2.5)$.
  • Directrix: $y = 3 - (-0.5) = 3.5$.

Identifying Conic Sections from General Form

When an equation is in the form $Ax^2 + Cy^2 + Dx + Ey + F = 0$:

1. Parabola: Only one squared term (either $x^2$ or $y^2$, but not both).
2. Hyperbola: Both squared terms are present and have opposite signs (one positive, one negative).
3. Circle: Both squared terms are present, have the same sign, and have equal coefficients ($A = C$).
4. Ellipse: Both squared terms are present, have the same sign, but have different coefficients ($A \neq C$).

Examples:

• $x^2 + 6x - 4y + 1 = 0 \rightarrow$ Parabola (only $x^2$ exists).
• $4x^2 - 9y^2 - 16x + 54y - 101 = 0 \rightarrow$ Hyperbola ($+4x^2$ vs $-9y^2$).
• $2x^2 + 8x + 2y^2 + 4y - 6 = 0 \rightarrow$ Circle ($2x^2$ and $2y^2$ coefficients match).
• $4x^2 + 25y^2 - 24x + 100y + 36 = 0 \rightarrow$ Ellipse ($4x^2$ and $25y^2$ are different signs, both positive).

Converting to Standard Form: Step-by-Step

Converting a Circle

Given: $2x^2 + 8x + 2y^2 + 4y - 6 = 0$

1. Group and move constant: $2(x^2 + 4x) + 2(y^2 + 2y) = 6$
2. Complete the square:
   • For $x$: $\left(\frac{4}{2}\right)^2 = 4$. Add to left side: $2(4) = 8$ added to right.
   • For $y$: $\left(\frac{2}{2}\right)^2 = 1$. Add to left side: $2(1) = 2$ added to right.
   • $2(x^2 + 4x + 4) + 2(y^2 + 2y + 1) = 6 + 8 + 2 = 16$
3. Factor: $2(x + 2)^2 + 2(y + 1)^2 = 16$
4. Divide: $(x + 2)^2 + (y + 1)^2 = 8$Result: Center $(-2, -1)$, radius $r = \sqrt{8} = 2\sqrt{2}$.

Converting an Ellipse

Given: $4x^2 + 25y^2 - 24x + 100y + 36 = 0$

1. Group: $(4x^2 - 24x) + (25y^2 + 100y) = -36$
2. Factor coefficients: $4(x^2 - 6x) + 25(y^2 + 4y) = -36$
3. Complete the square:
   • For $x$: $\left(\frac{-6}{2}\right)^2 = 9$. Add $4(9) = 36$ to right.
   • For $y$: $\left(\frac{4}{2}\right)^2 = 4$. Add $25(4) = 100$ to right.
   • $4(x^2 - 6x + 9) + 25(y^2 + 4y + 4) = -36 + 36 + 100 = 100$
4. Factor: $4(x - 3)^2 + 25(y + 2)^2 = 100$
5. Divide by 100: $\frac{(x - 3)^2}{25} + \frac{(y + 2)^2}{4} = 1$Result: Horizontal Ellipse, Center $(3, -2)$, $a = 5$, $b = 2$.

Converting a Hyperbola

Given: $4x^2 - 9y^2 - 16x + 54y - 101 = 0$

1. Group: $4(x^2 - 4x) - 9(y^2 - 6y) = 101$
2. Complete the square:
   • For $x$: $\left(\frac{-4}{2}\right)^2 = 4$. Add $4(4) = 16$ to right.
   • For $y$: $\left(\frac{-6}{2}\right)^2 = 9$. Add $-9(9) = -81$ to right.
   • $4(x^2 - 4x + 4) - 9(y^2 - 6y + 9) = 101 + 16 - 81 = 36$
3. Factor: $4(x - 2)^2 - 9(y - 3)^2 = 36$
4. Divide by 36: $\frac{(x - 2)^2}{9} - \frac{(y - 3)^2}{4} = 1$Result: Horizontal Hyperbola, Center $(2, 3)$, $a = 3, b = 2$.

Converting a Parabola

Given: $x^2 + 6x - 4y + 1 = 0$

1. Isolate squared variables: $x^2 + 6x = 4y - 1$
2. Complete the square: $\left(\frac{6}{2}\right)^2 = 9$. Add $9$ to both sides.
   • $x^2 + 6x + 9 = 4y - 1 + 9$
3. Factor: $(x + 3)^2 = 4y + 8$
4. Simplify: $(x + 3)^2 = 4(y + 2)$Result: Vertical Parabola, vertex $(-3, -2)$, $4p = 4 \rightarrow p = 1$.