Chapter 12 — Analytic Geometry
The conic sections — ellipse, hyperbola, and parabola — arise from slicing a cone with a plane. They appear everywhere: planetary orbits (ellipses), satellite dish reflectors (parabolas), cooling tower shapes (hyperbolas). This chapter develops the standard equations, key features, and graphs of each conic, along with rotation of axes and polar representations.
Table of Contents
1 — The Ellipse
- 1.1 Definition and Derivation
- 1.2 Standard Form (Center at Origin)
- 1.3 Translated Ellipses
- 1.4 Eccentricity
2 — The Hyperbola
- 2.1 Definition and Derivation
- 2.2 Standard Form (Center at Origin)
- 2.3 Asymptotes
- 2.4 Translated Hyperbolas
3 — The Parabola
- 3.1 Definition and Derivation
- 3.2 Standard Form (Vertex at Origin)
- 3.3 Translated Parabolas
- 3.4 Applications of Parabolas
4 — Rotation of Axes
- 4.1 The General Second-Degree Equation
- 4.2 Rotation Formulas
- 4.3 Identifying Conics Without Rotating
5 — Conic Sections in Polar Coordinates
Glossary
| Term | Definition |
|---|---|
| Conic section | Curve formed by intersecting a plane with a double cone |
| Focus (foci) | Fixed point(s) used in the definition of each conic |
| Directrix | Fixed line used in the definition of a parabola (or conic in polar form) |
| Eccentricity ($e$) | Ratio describing the shape: $e = 0$ circle, $0 < e < 1$ ellipse, $e = 1$ parabola, $e > 1$ hyperbola |
| Major axis | Longest axis of an ellipse; passes through both foci |
| Minor axis | Shortest axis of an ellipse; perpendicular to major axis |
| Transverse axis | Axis of a hyperbola that passes through the foci/vertices |
| Conjugate axis | Axis of a hyperbola perpendicular to the transverse axis |
| Vertex | Point(s) where a conic intersects its axis of symmetry |
| Latus rectum | Chord through a focus perpendicular to the axis |
1 — The Ellipse
1.1 Definition and Derivation
An ellipse is the set of all points in the plane such that the sum of the distances to two fixed points (foci) is constant:
\[d(P, F_1) + d(P, F_2) = 2a\]where $2a$ is the length of the major axis.
1.2 Standard Form (Center at Origin)
Horizontal major axis:
\[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \qquad (a > b > 0)\]| Feature | Value |
|---|---|
| Center | $(0, 0)$ |
| Vertices | $(\pm a, 0)$ |
| Co-vertices | $(0, \pm b)$ |
| Foci | $(\pm c, 0)$ where $c^2 = a^2 - b^2$ |
| Major axis length | $2a$ |
| Minor axis length | $2b$ |
Vertical major axis:
\[\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \qquad (a > b > 0)\]Vertices at $(0, \pm a)$, foci at $(0, \pm c)$.
Key relationship: $c^2 = a^2 - b^2$ (or equivalently $a^2 = b^2 + c^2$). The larger denominator is always $a^2$ and tells you the direction of the major axis.
Example: $\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1$
$a^2 = 25 \Rightarrow a = 5$, $b^2 = 9 \Rightarrow b = 3$, $c = \sqrt{25 - 9} = 4$
Vertices: $(\pm 5, 0)$. Foci: $(\pm 4, 0)$. Co-vertices: $(0, \pm 3)$.
1.3 Translated Ellipses
Center at $(h, k)$:
\[\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1\]All features shift by $(h, k)$.
Write the equation in standard form: $4x^2 + 9y^2 - 16x + 54y + 61 = 0$
Group and complete the square:
$4(x^2 - 4x) + 9(y^2 + 6y) = -61$
$4(x^2 - 4x + 4) + 9(y^2 + 6y + 9) = -61 + 16 + 81 = 36$
$\dfrac{(x - 2)^2}{9} + \dfrac{(y + 3)^2}{4} = 1$
Center: $(2, -3)$. $a = 3$ (horizontal), $b = 2$. $c = \sqrt{5}$.
1.4 Eccentricity
- $e = 0$: circle (foci coincide at center)
- $e$ close to $0$: nearly circular
- $e$ close to $1$: very elongated
Earth’s orbit: $e \approx 0.017$ (nearly circular). Pluto: $e \approx 0.25$.
2 — The Hyperbola
2.1 Definition and Derivation
A hyperbola is the set of all points such that the absolute difference of the distances to two foci is constant:
\(|d(P, F_1) - d(P, F_2)| = 2a\)
2.2 Standard Form (Center at Origin)
Horizontal transverse axis (opens left-right):
\[\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\]| Feature | Value |
|---|---|
| Center | $(0, 0)$ |
| Vertices | $(\pm a, 0)$ |
| Foci | $(\pm c, 0)$ where $c^2 = a^2 + b^2$ |
| Transverse axis length | $2a$ |
| Conjugate axis length | $2b$ |
Vertical transverse axis (opens up-down):
\(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\)
Hyperbola vs. Ellipse: For a hyperbola, $c^2 = a^2 \mathbf{+}\ b^2$ (note the $+$). For an ellipse, $c^2 = a^2 - b^2$. The positive term in the equation tells you the transverse axis direction.
2.3 Asymptotes
The hyperbola approaches (but never touches) two asymptotes:
Horizontal transverse axis: $y = \pm\dfrac{b}{a}x$
Vertical transverse axis: $y = \pm\dfrac{a}{b}x$
Graphing tip: Draw the central rectangle (corners at $(\pm a, \pm b)$), then draw the diagonals — these are the asymptotes.
2.4 Translated Hyperbolas
Center at $(h, k)$:
\[\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1\]Asymptotes: $y - k = \pm\dfrac{b}{a}(x - h)$
Example: $\dfrac{(x + 1)^2}{16} - \dfrac{(y - 2)^2}{9} = 1$
Center: $(-1, 2)$. $a = 4$, $b = 3$, $c = \sqrt{16 + 9} = 5$.
Vertices: $(-1 \pm 4, 2) = (-5, 2)$ and $(3, 2)$.
Foci: $(-1 \pm 5, 2) = (-6, 2)$ and $(4, 2)$.
Asymptotes: $y - 2 = \pm\dfrac{3}{4}(x + 1)$.
3 — The Parabola
3.1 Definition and Derivation
A parabola is the set of all points equidistant from a fixed point (focus) and a fixed line (directrix):
\(d(P, F) = d(P, \text{directrix})\)
3.2 Standard Form (Vertex at Origin)
| Equation | Opens | Focus | Directrix | Axis |
|---|---|---|---|---|
| $y^2 = 4px$ | Right ($p > 0$) | $(p, 0)$ | $x = -p$ | $y = 0$ |
| $y^2 = -4px$ | Left ($p > 0$) | $(-p, 0)$ | $x = p$ | $y = 0$ |
| $x^2 = 4py$ | Up ($p > 0$) | $(0, p)$ | $y = -p$ | $x = 0$ |
| $x^2 = -4py$ | Down ($p > 0$) | $(0, -p)$ | $y = p$ | $x = 0$ |
The parameter $p$ is the distance from vertex to focus (and from vertex to directrix).
Latus rectum length: $|4p|$ (width of parabola at the focus level).
Find focus, directrix, and latus rectum of $x^2 = 12y$.
$4p = 12 \Rightarrow p = 3$. Opens up.
Focus: $(0, 3)$. Directrix: $y = -3$. Latus rectum: $12$.
3.3 Translated Parabolas
Vertex at $(h, k)$:
\((y - k)^2 = 4p(x - h) \quad \text{(horizontal)}\) \((x - h)^2 = 4p(y - k) \quad \text{(vertical)}\)
Write the equation of the parabola with vertex $(2, -1)$, focus $(2, 3)$.
Opens up (focus above vertex). $p = 3 - (-1) = 4$.
$(x - 2)^2 = 4(4)(y + 1) = 16(y + 1)$
3.4 Applications of Parabolas
Reflective property: Signals arriving parallel to the axis reflect off a parabolic surface through the focus. This is why satellite dishes, telescopes, and car headlights use parabolic shapes.
Suspension bridges: Cables under uniform load hang in a parabola (not a catenary, which occurs under the cable’s own weight only).
4 — Rotation of Axes
4.1 The General Second-Degree Equation
When $B \ne 0$, the conic is rotated — its axes are not aligned with the coordinate axes.
4.2 Rotation Formulas
To eliminate the $xy$-term, rotate axes by angle $\theta$ where:
\[\cot 2\theta = \frac{A - C}{B}\]The rotation substitutions:
\(x = x'\cos\theta - y'\sin\theta\) \(y = x'\sin\theta + y'\cos\theta\)
In the new $(x’, y’)$ coordinates, the $B’x’y’$ term vanishes, revealing the standard conic.
4.3 Identifying Conics Without Rotating
The discriminant $B^2 - 4AC$ is invariant under rotation:
| Discriminant | Conic |
|---|---|
| $B^2 - 4AC < 0$ | Ellipse (or circle if $A = C$ and $B = 0$) |
| $B^2 - 4AC = 0$ | Parabola |
| $B^2 - 4AC > 0$ | Hyperbola |
(Degenerate cases — point, line, two lines — are possible.)
Identify: $3x^2 + 2xy + 3y^2 - 8 = 0$.
$A = 3$, $B = 2$, $C = 3$.
$B^2 - 4AC = 4 - 36 = -32 < 0$ → Ellipse
5 — Conic Sections in Polar Coordinates
5.1 Focus-Directrix Definition
A conic with eccentricity $e$ and directrix at distance $d$ from the focus satisfies:
\[\frac{d(P, F)}{d(P, \ell)} = e\]- $e < 1$: ellipse
- $e = 1$: parabola
- $e > 1$: hyperbola
5.2 Polar Equations of Conics
With one focus at the pole:
| Directrix | Polar Equation | |———–|—————| | $x = d$ (right of pole) | $r = \dfrac{ed}{1 + e\cos\theta}$ | | $x = -d$ (left of pole) | $r = \dfrac{ed}{1 - e\cos\theta}$ | | $y = d$ (above pole) | $r = \dfrac{ed}{1 + e\sin\theta}$ | | $y = -d$ (below pole) | $r = \dfrac{ed}{1 - e\sin\theta}$ |
Identify and describe: $r = \dfrac{6}{2 + \cos\theta}$.
Divide numerator and denominator by $2$: $r = \dfrac{3}{1 + \frac{1}{2}\cos\theta}$
$e = \dfrac{1}{2} < 1$ → Ellipse. $ed = 3 \Rightarrow d = 6$.
Semi-major axis can be found from the vertices (set $\theta = 0$ and $\theta = \pi$):
$r(0) = \dfrac{6}{3} = 2$, $r(\pi) = \dfrac{6}{1} = 6$.
Major axis $= 2 + 6 = 8$, so $a = 4$.
Key Takeaways
- Ellipse: Sum of distances to foci is constant ($2a$). $c^2 = a^2 - b^2$. Larger denominator = major axis direction.
- Hyperbola: Absolute difference of distances to foci is constant ($2a$). $c^2 = a^2 + b^2$. Asymptotes from the central rectangle.
- Parabola: Equal distance from focus and directrix. $p$ = vertex-to-focus distance. Reflective property.
- Completing the square converts general form to standard form for ellipses and hyperbolas.
- Discriminant $B^2 - 4AC$ classifies any general second-degree equation without rotating.
- Rotation of axes eliminates the $xy$-term when the conic is tilted.
- Polar conics place one focus at the origin; eccentricity $e$ determines the type.
- Eccentricity spectrum: $e = 0$ circle → $0 < e < 1$ ellipse → $e = 1$ parabola → $e > 1$ hyperbola.
Practice Questions
Q1. Find the foci and vertices of $\dfrac{x^2}{36} + \dfrac{y^2}{100} = 1$.
Answer:
$a^2 = 100$ (under $y^2$, so vertical major axis), $b^2 = 36$, $c^2 = 100 - 36 = 64 \Rightarrow c = 8$.
Vertices: $(0, \pm 10)$. Foci: $(0, \pm 8)$.
Q2. Find the equation of the ellipse with foci $(\pm 3, 0)$ and vertices $(\pm 5, 0)$.
Answer:
$a = 5$, $c = 3$. $b^2 = 25 - 9 = 16$.
\(\frac{x^2}{25} + \frac{y^2}{16} = 1\)
Q3. Find the asymptotes and foci of $\dfrac{y^2}{4} - \dfrac{x^2}{9} = 1$.
Answer:
Vertical transverse axis. $a = 2$, $b = 3$, $c = \sqrt{4 + 9} = \sqrt{13}$.
Asymptotes: $y = \pm\dfrac{2}{3}x$. Foci: $(0, \pm\sqrt{13})$.
Q4. Find the focus and directrix of $y^2 = -20x$.
Answer:
$4p = 20 \Rightarrow p = 5$. Opens left (negative coefficient).
Focus: $(-5, 0)$. Directrix: $x = 5$.
Q5. Classify without rotating: $2x^2 + 3xy - 4y^2 + 7 = 0$.
Answer:
$A = 2$, $B = 3$, $C = -4$.
$B^2 - 4AC = 9 - 4(2)(-4) = 9 + 32 = 41 > 0$ → Hyperbola
Q6. Identify and find eccentricity: $r = \dfrac{12}{3 - 3\sin\theta}$.
Answer:
Divide by $3$: $r = \dfrac{4}{1 - \sin\theta}$.
$e = 1$ → Parabola. $ed = 4 \Rightarrow d = 4$.
Directrix: $y = -4$ (below pole). Focus at the pole.