6.2 Conic Sections: Parabolas, Ellipses, and Hyperbolas

6.2 Conic Sections: Parabolas, Ellipses, and Hyperbolas

Conic sections are curves generated by intersecting a double-napped right circular cone with a plane. Depending on the angle of intersection relative to the generator line of the cone, the resulting non-degenerate curves are classified as parabolas, ellipses (including circles), or hyperbolas.

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1. General Equation and Conic Classification

Without rotation of axes, the general equation of a second-degree polynomial in two variables is:

$Ax^2 + Cy^2 + Dx + Ey + F = 0$

Assuming the equation represents a non-degenerate conic section, the conic type is determined by comparing the coefficients $A$ and $C$ of the squared terms:

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2. Parabolas: Locus, Focus, Directrix, and Latus Rectum

A parabola is the locus of all points $P(x, y)$ equidistant from a fixed point $F$ (the focus) and a fixed line $D$ (the directrix). The point halfway between the focus and directrix is the vertex $(h, k)$. The parameter $p$ represents the directed distance from the vertex to the focus.

Standard Forms of Parabolas

  Vertical Axis: (x - h)^2 = 4p(y - k)          Horizontal Axis: (y - k)^2 = 4p(x - h)
           Opens UP if p > 0                             Opens RIGHT if p > 0
          Opens DOWN if p < 0                             Opens LEFT if p < 0

• Vertical Axis of Symmetry ($x = h$):

  • Standard Equation: $(x - h)^2 = 4p(y - k)$
  • Focus: $(h, k + p)$
  • Directrix Line: $y = k - p$
  • Focal Length: $|p|$; Focal Width (Latus Rectum): $|4p|$

• Horizontal Axis of Symmetry ($y = k$):

  • Standard Equation: $(y - k)^2 = 4p(x - h)$
  • Focus: $(h + p, k)$
  • Directrix Line: $x = h - p$
  • Focal Length: $|p|$; Focal Width (Latus Rectum): $|4p|$

Worked Example 1: Parabola Analysis via Completing the Square

Determine the vertex, focus, directrix, and axis of symmetry for the parabola:

$y^2 + 6y + 8x - 7 = 0$

Solution:

1. Isolate $y$-terms on the left side and $x$-terms/constants on the right side: $y^2 + 6y = -8x + 7$
2. Complete the square for $y$ by adding $\left(\frac{6}{2}\right)^2 = 9$ to both sides: $y^2 + 6y + 9 = -8x + 7 + 9 \implies (y + 3)^2 = -8x + 16$
3. Factor out the coefficient of $x$ on the right side to achieve standard form: $(y + 3)^2 = -8(x - 2)$
4. Extract key parabolic parameters:
   • Standard form: $(y - k)^2 = 4p(x - h) \implies (h, k) = (2, -3)$
   • Focal parameter: $4p = -8 \implies p = -2$ (opens left)
   • Focus: $(h + p, k) = (2 + (-2), -3) = (0, -3)$
   • Directrix Line: $x = h - p = 2 - (-2) \implies x = 4$
   • Axis of Symmetry: $y = -3$

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3. Ellipses: Axes, Foci, and Eccentricity

An ellipse is the locus of points $P(x, y)$ such that the sum of distances from $P$ to two fixed foci $F_1$ and $F_2$ is a constant $2a$. The center of the ellipse is $(h, k)$, $a$ is the semi-major axis length, $b$ is the semi-minor axis length, and $c$ is the focal distance, satisfying $a > b > 0$ and $a^2 = b^2 + c^2$.

Standard Equations of Ellipses

Worked Example 2: Ellipse Equation and Focal Extraction

Convert $9x^2 + 25y^2 - 36x + 50y - 164 = 0$ into standard form. Find the center, vertices, foci, and eccentricity.

Solution:

1. Group variables and factor leading coefficients: $9(x^2 - 4x) + 25(y^2 + 2y) = 164$
2. Complete the squares inside parentheses: $9(x^2 - 4x + 4) + 25(y^2 + 2y + 1) = 164 + 9(4) + 25(1)$ $9(x - 2)^2 + 25(y + 1)^2 = 164 + 36 + 25 = 225$
3. Divide by $225$ to obtain standard form: $\frac{9(x - 2)^2}{225} + \frac{25(y + 1)^2}{225} = 1 \implies \frac{(x - 2)^2}{25} + \frac{(y + 1)^2}{9} = 1$
4. Identify parameters:
   • Center: $(h, k) = (2, -1)$
   • Major axis horizontal ($a^2 = 25 \implies a = 5$, $b^2 = 9 \implies b = 3$)
   • Focal distance: $c = \sqrt{a^2 - b^2} = \sqrt{25 - 9} = \sqrt{16} = 4$
   • Major Vertices: $(h \pm a, k) = (2 \pm 5, -1) \implies (7, -1)$ and $(-3, -1)$
   • Foci: $(h \pm c, k) = (2 \pm 4, -1) \implies (6, -1)$ and $(-2, -1)$
   • Eccentricity: $e = \frac{c}{a} = \frac{4}{5} = 0.8$

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4. Hyperbolas: Transverse Axes, Foci, and Asymptotes

A hyperbola is the locus of points $P(x, y)$ such that the absolute difference of distances to two foci $F_1$ and $F_2$ is a constant $2a$. The focal distance $c$ satisfies $c^2 = a^2 + b^2$, meaning $c > a$.

Standard Equations of Hyperbolas

Worked Example 3: Hyperbola Analysis with Asymptote Equations

Convert $16x^2 - 9y^2 - 64x - 54y - 161 = 0$ into standard form. Find the center, vertices, foci, and asymptotes.

Solution:

1. Group terms and factor out coefficients: $16(x^2 - 4x) - 9(y^2 + 6y) = 161$
2. Complete the squares (note the negative factor $-9$ distributed on $y$): $16(x^2 - 4x + 4) - 9(y^2 + 6y + 9) = 161 + 16(4) - 9(9)$ $16(x - 2)^2 - 9(y + 3)^2 = 161 + 64 - 81 = 144$
3. Divide by $144$: $\frac{(x - 2)^2}{9} - \frac{(y + 3)^2}{16} = 1$
4. Compute parameters:
   • Center: $(h, k) = (2, -3)$
   • Transverse axis is horizontal ($a^2 = 9 \implies a = 3$, $b^2 = 16 \implies b = 4$)
   • Focal length: $c = \sqrt{a^2 + b^2} = \sqrt{9 + 16} = 5$
   • Vertices: $(2 \pm 3, -3) \implies (5, -3)$ and $(-1, -3)$
   • Foci: $(2 \pm 5, -3) \implies (7, -3)$ and $(-3, -3)$
   • Asymptote Equations: $y - (-3) = \pm \frac{4}{3}(x - 2) \implies y + 3 = \pm \frac{4}{3}(x - 2)$

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5. CLEP Exam Traps & Common Errors

• Trap 1: Confusing Focal Relations ($a^2 = b^2 + c^2$ vs. $c^2 = a^2 + b^2$)
  For ellipses, $a$ is the longest semi-axis, so $a^2 = b^2 + c^2 \implies c = \sqrt{a^2 - b^2}$. For hyperbolas, $c$ is the longest parameter from the center to the foci, so $c^2 = a^2 + b^2 \implies c = \sqrt{a^2 + b^2}$. Using the wrong formula flips focal locations.
• Trap 2: Asymptote Slope Numerator/Denominator Inversion
  Remember that slope is always $\frac{\Delta y}{\Delta x}$. The parameter under the $y$-term in the standard equation (square-rooted) is always the numerator of the asymptote slope, while the parameter under the $x$-term is the denominator.
• Trap 3: Sign Errors when Completing the Square on Hyperbolas
  When factoring $-9(y^2 + 6y)$, completing the square requires adding $+9$ inside the parenthesis. Adding $+9$ inside a parenthesis scaled by $-9$ actually subtracts $81$ from the left side, so you must subtract $81$ from the right side as well.