Conic Sections: A Complete Tutorial

A comprehensive journey through circles, ellipses, parabolas, and hyperbolas—from foundational concepts to advanced applications.

Formula Quick Reference (click to expand)
Circle \((x-h)^2 + (y-k)^2 = r^2\)
Center: \((h,k)\), Radius: \(r\)
Ellipse (horizontal) \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\)
\(c^2 = a^2 - b^2\), \(e = c/a \lt 1\)
Parabola (vertical) \((x-h)^2 = 4p(y-k)\)
Focus: \((h, k+p)\), Directrix: \(y = k-p\)
Hyperbola (horizontal) \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\)
\(c^2 = a^2 + b^2\), \(e = c/a \gt 1\)
Discriminant Test For \(Ax^2 + Bxy + Cy^2 + ... = 0\):
\(B^2 - 4AC \lt 0\): Ellipse
\(B^2 - 4AC = 0\): Parabola
\(B^2 - 4AC \gt 0\): Hyperbola
Polar Form \(r = \frac{ed}{1 \pm e\cos\theta}\)
\(e\) = eccentricity, \(d\) = focus-directrix distance
FOUNDATION LEVEL

Foundation: Understanding Conic Sections

What Are Conic Sections?

Conic sections are curves obtained by intersecting a plane with a double cone (two identical cones placed vertex-to-vertex). The type of conic section depends on the angle at which the plane intersects the cone relative to the cone's axis.

Historical Context
The ancient Greek mathematician Apollonius of Perga (c. 262-190 BCE) wrote the definitive treatise on conic sections, "Conics," comprising eight books. He gave the ellipse, parabola, and hyperbola their names and developed their fundamental properties. His work was so comprehensive that it remained the standard reference for over 1,800 years.

The Focus-Directrix Definition

A more unified definition uses the concept of eccentricity (\(e\)):

Definition: A conic section is the locus of all points \(P\) such that the ratio of the distance from \(P\) to a fixed point (the focus \(F\)) to the distance from \(P\) to a fixed line (the directrix \(d\)) is a constant \(e\), called the eccentricity.
$$\frac{\text{distance from } P \text{ to focus}}{\text{distance from } P \text{ to directrix}} = e$$

The value of eccentricity determines the type of conic:

Conic Type Eccentricity (\(e\)) Geometric Property
Circle \(e = 0\) All points equidistant from center
Ellipse \(0 \lt e \lt 1\) Sum of distances to two foci is constant
Parabola \(e = 1\) Equal distance to focus and directrix
Hyperbola \(e \gt 1\) Absolute difference of distances to two foci is constant
Unifying Role of Eccentricity: Eccentricity measures how "stretched" a conic is from being circular. As \(e\) increases from 0, circles become ellipses, then at \(e = 1\) we have parabolas (which are "infinitely stretched"), and beyond that hyperbolas which open up completely.

Interactive: Eccentricity Morphing

Drag the slider to see how the conic section transforms as eccentricity changes from 0 to 2.

e = 0.00 (Circle)
FOUNDATION LEVEL

Circle (\(e = 0\))

Definition and Properties

A circle is the set of all points in a plane that are equidistant from a fixed point called the center. This constant distance is called the radius (\(r\)).

Standard Form (Center at origin): $$x^2 + y^2 = r^2$$
Standard Form (Center at \((h, k)\)): $$(x - h)^2 + (y - k)^2 = r^2$$

Key Features

Example: Circle Equations (Both Directions)

These two problems show how the same circle can be expressed in different forms.

Part A - Standard Form from Center and Radius: Find the equation given center \((3, -2)\) and radius \(5\).

Use the standard form: \((x - h)^2 + (y - k)^2 = r^2\)
Substitute \(h = 3\), \(k = -2\), and \(r = 5\)
$$(x - 3)^2 + (y + 2)^2 = 25$$

Part B - Center and Radius from General Form: Convert \(x^2 + y^2 - 6x + 4y - 12 = 0\) to standard form.

Group terms: \((x^2 - 6x) + (y^2 + 4y) = 12\)
Complete the square: add \((\frac{-6}{2})^2 = 9\) and \((\frac{4}{2})^2 = 4\) to both sides
$$(x - 3)^2 + (y + 2)^2 = 25$$

Result: Center \((3, -2)\), Radius \(5\) — the same circle as Part A!

Practice Problems: Circle

  1. Find the equation of a circle with center \((-4, 5)\) and radius \(7\).
  2. Convert \(x^2 + y^2 + 8x - 10y + 16 = 0\) to standard form.
  3. Find the center and radius of \(x^2 + y^2 - 2x + 6y - 15 = 0\).
  4. Determine if the point \((2, 3)\) lies inside, outside, or on the circle \((x - 1)^2 + (y - 1)^2 = 9\).
Show Solutions

Solution 1: \((x + 4)^2 + (y - 5)^2 = 49\)

Solution 2:

\((x^2 + 8x) + (y^2 - 10y) = -16\)
\((x^2 + 8x + 16) + (y^2 - 10y + 25) = -16 + 16 + 25\)
\((x + 4)^2 + (y - 5)^2 = 25\)

Solution 3:

\((x^2 - 2x + 1) + (y^2 + 6y + 9) = 15 + 1 + 9\)
\((x - 1)^2 + (y + 3)^2 = 25\)

Center: \((1, -3)\), Radius: \(5\)

Solution 4:

Distance from \((2, 3)\) to center \((1, 1)\): \(d = \sqrt{(2-1)^2 + (3-1)^2} = \sqrt{5} \approx 2.24\)
Radius: \(r = 3\)

Since \(d \lt r\), the point is inside the circle.

FOUNDATION LEVEL

Ellipse (\(0 \lt e \lt 1\))

Definition and Properties

An ellipse is the set of all points in a plane such that the sum of the distances from two fixed points (called foci) is constant.

Focal Definition: For any point \(P\) on the ellipse with foci \(F_1\) and \(F_2\): $$|PF_1| + |PF_2| = 2a$$ where \(a\) is the length of the semi-major axis.

Standard Forms

Horizontal Major Axis (Center at origin): $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ where \(a \gt b\)
Vertical Major Axis (Center at origin): $$\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$$ where \(a \gt b\)
General Form (Center at \((h, k)\)): $$\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \quad \text{(horizontal)}$$ $$\frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1 \quad \text{(vertical)}$$

Key Features and Formulas

Why does \(c^2 = a^2 - b^2\)? (Derivation)

Consider a point \(P\) at the co-vertex \((0, b)\) on the ellipse. By the focal definition, the sum of distances to both foci equals \(2a\).

The foci are at \((-c, 0)\) and \((c, 0)\)
Distance from \((0, b)\) to either focus: \(\sqrt{c^2 + b^2}\)
Since both distances are equal: \(2\sqrt{c^2 + b^2} = 2a\)
Simplify: \(\sqrt{c^2 + b^2} = a\)
Square both sides: \(c^2 + b^2 = a^2\)
Rearrange: \(c^2 = a^2 - b^2\)
Don't confuse \(a\) and \(b\)! The value under the larger denominator determines the direction of the major axis, not the variable name. Always check which is larger.

Interactive: Focus-Directrix Relationship

Drag point P around the ellipse to see that the ratio of (distance to focus) / (distance to directrix) always equals eccentricity e.

e = 0.50

Example 3: Finding the Equation of an Ellipse

Problem: Find the equation of an ellipse with center at origin, vertices at \((\pm 5, 0)\), and co-vertices at \((0, \pm 3)\).

Solution:

Vertices at \((\pm 5, 0)\) indicate horizontal major axis with \(a = 5\)
Co-vertices at \((0, \pm 3)\) indicate \(b = 3\)
Use standard form for horizontal ellipse:
$$\frac{x^2}{25} + \frac{y^2}{9} = 1$$
Find focal distance: \(c^2 = a^2 - b^2 = 25 - 9 = 16\), so \(c = 4\)
Foci are at \((\pm 4, 0)\)
Eccentricity: \(e = \frac{c}{a} = \frac{4}{5} = 0.8\)

Example 4: Converting General Form to Standard Form

Problem: Convert \(9x^2 + 4y^2 - 36x + 8y + 4 = 0\) to standard form and identify all key features.

Solution:

Group terms: \(9(x^2 - 4x) + 4(y^2 + 2y) = -4\)
Complete the square for \(x\): \(x^2 - 4x + 4\) (add \(9 \cdot 4 = 36\) to both sides)
Complete the square for \(y\): \(y^2 + 2y + 1\) (add \(4 \cdot 1 = 4\) to both sides)
$$9(x^2 - 4x + 4) + 4(y^2 + 2y + 1) = -4 + 36 + 4$$ $$9(x - 2)^2 + 4(y + 1)^2 = 36$$
Divide by 36 to get standard form:
$$\frac{(x - 2)^2}{4} + \frac{(y + 1)^2}{9} = 1$$

Key Features:

  • Center: \((2, -1)\)
  • Vertical major axis (since \(9 \gt 4\))
  • \(a^2 = 9\), so \(a = 3\); \(b^2 = 4\), so \(b = 2\)
  • \(c^2 = 9 - 4 = 5\), so \(c = \sqrt{5}\)
  • Vertices: \((2, -1 \pm 3) = (2, 2)\) and \((2, -4)\)
  • Co-vertices: \((2 \pm 2, -1) = (4, -1)\) and \((0, -1)\)
  • Foci: \((2, -1 \pm \sqrt{5})\)
  • Eccentricity: \(e = \frac{\sqrt{5}}{3} \approx 0.745\)

Example 5: Using the Focal Definition

Problem: An ellipse has foci at \((0, 3)\) and \((0, -3)\), and the sum of distances from any point on the ellipse to the foci is 10. Find the equation.

Solution:

Foci on \(y\)-axis indicate vertical major axis
Sum of distances = \(2a = 10\), so \(a = 5\)
Foci at \((0, \pm 3)\) means \(c = 3\)
Find \(b\): \(c^2 = a^2 - b^2\)
$$9 = 25 - b^2$$ $$b^2 = 16$$ $$b = 4$$
Standard form (vertical):
$$\frac{x^2}{16} + \frac{y^2}{25} = 1$$

Practice Problems: Ellipse

  1. Find the equation of an ellipse with center \((-2, 3)\), horizontal major axis of length 12, and minor axis of length 8.
  2. Convert \(25x^2 + 9y^2 = 225\) to standard form and find the foci.
  3. An ellipse has vertices at \((1, 7)\) and \((1, -1)\), and co-vertices at \((-2, 3)\) and \((4, 3)\). Find its equation.
  4. Find the eccentricity of the ellipse \(\frac{x^2}{64} + \frac{y^2}{100} = 1\).
  5. Convert \(4x^2 + y^2 - 8x + 4y - 8 = 0\) to standard form.
Show Solutions

Solution 1:

Major axis length = 12, so \(2a = 12\) and \(a = 6\)
Minor axis length = 8, so \(2b = 8\) and \(b = 4\)
Equation: \(\frac{(x + 2)^2}{36} + \frac{(y - 3)^2}{16} = 1\)

Solution 2:

Divide by 225: \(\frac{x^2}{9} + \frac{y^2}{25} = 1\)
\(a^2 = 25\) (vertical), \(b^2 = 9\)
\(c^2 = 25 - 9 = 16\), so \(c = 4\)
Foci: \((0, \pm 4)\)

Solution 3:

Center (midpoint of vertices): \((1, 3)\)
Distance from center to vertex: \(a = 4\) (vertical)
Distance from center to co-vertex: \(b = 3\)
Equation: \(\frac{(x - 1)^2}{9} + \frac{(y - 3)^2}{16} = 1\)

Solution 4:

\(a^2 = 100\) (vertical), so \(a = 10\); \(b^2 = 64\), so \(b = 8\)
\(c^2 = 100 - 64 = 36\), so \(c = 6\)
\(e = \frac{c}{a} = \frac{6}{10} = 0.6\)

Solution 5:

\(4(x^2 - 2x) + (y^2 + 4y) = 8\)
\(4(x^2 - 2x + 1) + (y^2 + 4y + 4) = 8 + 4 + 4\)
\(4(x - 1)^2 + (y + 2)^2 = 16\)
\(\frac{(x - 1)^2}{4} + \frac{(y + 2)^2}{16} = 1\)
Historical Context
Johannes Kepler (1571-1630) discovered that planetary orbits are ellipses with the Sun at one focus, not circles as previously believed. This First Law of Planetary Motion revolutionized astronomy and was one of the key discoveries that led to Newton's law of universal gravitation.
FOUNDATION LEVEL

Parabola (\(e = 1\))

Definition and Properties

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

Focal-Directrix Definition: For any point \(P\) on the parabola: $$\text{distance from } P \text{ to focus} = \text{distance from } P \text{ to directrix}$$

Standard Forms

Vertical Axis (Opens up/down): $$x^2 = 4py$$
  • Vertex at origin
  • Opens up if \(p \gt 0\)
  • Opens down if \(p \lt 0\)
  • Focus: \((0, p)\)
  • Directrix: \(y = -p\)
Horizontal Axis (Opens left/right): $$y^2 = 4px$$
  • Vertex at origin
  • Opens right if \(p \gt 0\)
  • Opens left if \(p \lt 0\)
  • Focus: \((p, 0)\)
  • Directrix: \(x = -p\)
General Forms (Vertex at \((h, k)\)): $$(x - h)^2 = 4p(y - k) \quad \text{(vertical)}$$ $$(y - k)^2 = 4p(x - h) \quad \text{(horizontal)}$$

Key Features

Example 6: Finding Parabola from Focus and Directrix

Problem: Find the equation of a parabola with focus at \((0, 3)\) and directrix \(y = -3\).

Solution:

Vertex is midpoint between focus and directrix: \((0, 0)\)
Distance from vertex to focus: \(p = 3\)
Since focus is above vertex, parabola opens upward
Use vertical form: \(x^2 = 4py\)
$$x^2 = 4(3)y$$ $$x^2 = 12y$$

Example 7: Converting to Standard Form

Problem: Convert \(y^2 - 6y - 8x + 1 = 0\) to standard form and identify all key features.

Solution:

Isolate terms with \(y\): \(y^2 - 6y = 8x - 1\)
Complete the square: \(y^2 - 6y + 9 = 8x - 1 + 9\)
$$(y - 3)^2 = 8x + 8$$ $$(y - 3)^2 = 8(x + 1)$$
Compare with \((y - k)^2 = 4p(x - h)\): \(4p = 8\), so \(p = 2\)

Key Features:

  • Vertex: \((-1, 3)\)
  • Opens right (horizontal, \(p \gt 0\))
  • Focus: \((-1 + 2, 3) = (1, 3)\)
  • Directrix: \(x = -1 - 2 = -3\)
  • Axis of symmetry: \(y = 3\)
  • Latus rectum length: \(|4p| = 8\)

Practice Problems: Parabola

  1. Find the equation of a parabola with vertex \((2, -1)\) and focus \((2, 3)\).
  2. Convert \(x^2 + 4x - 12y + 16 = 0\) to standard form and find the focus and directrix.
  3. A parabola has focus \((-3, 0)\) and directrix \(x = 3\). Find its equation.
  4. Find the vertex and focus of \(y^2 = -16x\).
Show Solutions

Solution 1:

Vertex at \((2, -1)\), focus at \((2, 3)\)
\(p = 3 - (-1) = 4\) (vertical, upward)
\((x - 2)^2 = 4(4)(y + 1) = 16(y + 1)\)

Solution 2:

\(x^2 + 4x = 12y - 16\)
\(x^2 + 4x + 4 = 12y - 16 + 4\)
\((x + 2)^2 = 12y - 12 = 12(y - 1)\)
\(4p = 12\), so \(p = 3\)
Vertex: \((-2, 1)\), Focus: \((-2, 4)\), Directrix: \(y = -2\)

Solution 3:

Vertex is midpoint: \((0, 0)\)
\(p = -3\) (opens left)
\(y^2 = 4(-3)x = -12x\)

Solution 4:

\(4p = -16\), so \(p = -4\)
Vertex: \((0, 0)\), Focus: \((-4, 0)\)
Historical Context
Galileo Galilei (1564-1642) discovered that projectiles follow parabolic paths. This was groundbreaking because it showed that the same mathematical laws govern both earthly and celestial motion, contradicting Aristotelian physics which claimed different rules for each realm.
FOUNDATION LEVEL

Hyperbola (\(e \gt 1\))

Definition and Properties

A hyperbola is the set of all points in a plane such that the absolute value of the difference of the distances from two fixed points (the foci) is constant.

Focal Definition: For any point \(P\) on the hyperbola with foci \(F_1\) and \(F_2\): $$||PF_1| - |PF_2|| = 2a$$ where \(a\) is the distance from center to vertex.

Standard Forms

Horizontal Transverse Axis: $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$
  • Opens left and right
  • Vertices: \((\pm a, 0)\)
  • Foci: \((\pm c, 0)\)
Vertical Transverse Axis: $$\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$$
  • Opens up and down
  • Vertices: \((0, \pm a)\)
  • Foci: \((0, \pm c)\)
General Forms (Center at \((h, k)\)): $$\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \quad \text{(horizontal)}$$ $$\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \quad \text{(vertical)}$$

Key Features and Formulas

Unlike the ellipse, for a hyperbola \(c^2 = a^2 + b^2\) (note the plus sign!). Also, \(a\) is NOT necessarily larger than \(b\).

Example 8: Finding Hyperbola from Vertices and Foci

Problem: Find the equation of a hyperbola with center at origin, vertices at \((0, \pm 4)\), and foci at \((0, \pm 5)\).

Solution:

Vertices on \(y\)-axis indicate vertical transverse axis
\(a = 4\) (distance from center to vertex)
\(c = 5\) (distance from center to focus)
Find \(b\): \(c^2 = a^2 + b^2\)
$$25 = 16 + b^2$$ $$b^2 = 9$$ $$b = 3$$
Standard form (vertical):
$$\frac{y^2}{16} - \frac{x^2}{9} = 1$$
Asymptotes: \(y = \pm \frac{4}{3}x\)
Eccentricity: \(e = \frac{5}{4} = 1.25\)

Example 9: Converting General Form to Standard Form

Problem: Convert \(9x^2 - 16y^2 - 54x - 32y - 79 = 0\) to standard form and identify all key features.

Solution:

Group terms: \(9(x^2 - 6x) - 16(y^2 + 2y) = 79\)
Complete the square for \(x\): \(x^2 - 6x + 9\) (add \(9 \cdot 9 = 81\))
Complete the square for \(y\): \(y^2 + 2y + 1\) (subtract \(16 \cdot 1 = 16\))
$$9(x^2 - 6x + 9) - 16(y^2 + 2y + 1) = 79 + 81 - 16$$ $$9(x - 3)^2 - 16(y + 1)^2 = 144$$
Divide by 144:
$$\frac{(x - 3)^2}{16} - \frac{(y + 1)^2}{9} = 1$$

Key Features:

  • Center: \((3, -1)\)
  • Horizontal transverse axis
  • \(a^2 = 16\), so \(a = 4\); \(b^2 = 9\), so \(b = 3\)
  • \(c^2 = 16 + 9 = 25\), so \(c = 5\)
  • Vertices: \((3 \pm 4, -1) = (7, -1)\) and \((-1, -1)\)
  • Foci: \((3 \pm 5, -1) = (8, -1)\) and \((-2, -1)\)
  • Asymptotes: \(y + 1 = \pm \frac{3}{4}(x - 3)\)
  • Eccentricity: \(e = \frac{5}{4} = 1.25\)

Practice Problems: Hyperbola

  1. Find the equation of a hyperbola with center \((0, 0)\), vertices \((\pm 3, 0)\), and foci \((\pm 5, 0)\).
  2. Convert \(4y^2 - 9x^2 = 36\) to standard form and find the vertices and foci.
  3. Find the asymptotes of \(\frac{x^2}{16} - \frac{y^2}{25} = 1\).
  4. A hyperbola has foci at \((0, -6)\) and \((0, 6)\), and vertices at \((0, -4)\) and \((0, 4)\). Find its equation and eccentricity.
Show Solutions

Solution 1:

Horizontal: \(a = 3\), \(c = 5\)
\(b^2 = c^2 - a^2 = 25 - 9 = 16\)
Equation: \(\frac{x^2}{9} - \frac{y^2}{16} = 1\)

Solution 2:

Divide by 36: \(\frac{y^2}{9} - \frac{x^2}{4} = 1\)
Vertical: \(a^2 = 9\), so \(a = 3\); \(b^2 = 4\), so \(b = 2\)
\(c^2 = 9 + 4 = 13\), so \(c = \sqrt{13}\)
Vertices: \((0, \pm 3)\), Foci: \((0, \pm \sqrt{13})\)

Solution 3:

Horizontal: \(a = 4\), \(b = 5\)
Asymptotes: \(y = \pm \frac{5}{4}x\)

Solution 4:

Vertical: \(a = 4\), \(c = 6\)
\(b^2 = 36 - 16 = 20\)
Equation: \(\frac{y^2}{16} - \frac{x^2}{20} = 1\)
Eccentricity: \(e = \frac{6}{4} = 1.5\)
INTERMEDIATE LEVEL

Intermediate Techniques

Identifying Conics Using the Discriminant

The general second-degree equation in two variables is:

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

The type of conic can be determined using the discriminant \(B^2 - 4AC\):

Discriminant Conic Type Additional Conditions
\(B^2 - 4AC \lt 0\) Ellipse If \(A = C\) and \(B = 0\), it's a circle
\(B^2 - 4AC = 0\) Parabola —
\(B^2 - 4AC \gt 0\) Hyperbola —
Any Degenerate case If equation factors into lines or a point

Example 10: Identifying Conics

Problem: Identify the conic represented by each equation:

(a) \(3x^2 + 2xy + 3y^2 - 4x + 2y - 5 = 0\)

(b) \(x^2 - 4xy + 4y^2 + 2x - y + 1 = 0\)

(c) \(2x^2 - 3xy - 2y^2 + x + 3y - 1 = 0\)

Solution:

(a) \(A = 3\), \(B = 2\), \(C = 3\)

$$B^2 - 4AC = 4 - 4(3)(3) = 4 - 36 = -32 \lt 0$$

This is an ellipse.

(b) \(A = 1\), \(B = -4\), \(C = 4\)

$$B^2 - 4AC = 16 - 4(1)(4) = 0$$

This is a parabola.

(c) \(A = 2\), \(B = -3\), \(C = -2\)

$$B^2 - 4AC = 9 - 4(2)(-2) = 25 \gt 0$$

This is a hyperbola.

Completing the Square

To convert a general second-degree equation (without the \(xy\) term) to standard form, we use the method of completing the square.

Steps for Completing the Square:
  1. Group \(x\) and \(y\) terms separately
  2. Factor out coefficients of \(x^2\) and \(y^2\)
  3. Complete the square for each variable
  4. Add appropriate values to both sides
  5. Convert to standard form

Example 11: Complete the Square for an Ellipse

Problem: Convert \(4x^2 + 9y^2 - 16x + 18y - 11 = 0\) to standard form.

Solution:

Group and factor: \(4(x^2 - 4x) + 9(y^2 + 2y) = 11\)
Complete square for \(x\): \(x^2 - 4x + 4\) (add \(4 \cdot 4 = 16\) to right)
Complete square for \(y\): \(y^2 + 2y + 1\) (add \(9 \cdot 1 = 9\) to right)
$$4(x - 2)^2 + 9(y + 1)^2 = 11 + 16 + 9 = 36$$
Divide by 36:
$$\frac{(x - 2)^2}{9} + \frac{(y + 1)^2}{4} = 1$$

This is an ellipse with center \((2, -1)\), \(a = 3\) (horizontal), \(b = 2\).

Tangent Lines and Normal Lines

For a conic section, the tangent line at a point touches the curve at exactly one point (locally), while the normal line is perpendicular to the tangent at that point.

Finding Tangent Lines:
  1. Use implicit differentiation to find \(\frac{dy}{dx}\)
  2. Evaluate at the point of tangency to get slope
  3. Use point-slope form: \(y - y_1 = m(x - x_1)\)
Alternative Tangent Formulas:

For ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), the tangent at \((x_0, y_0)\) is:

$$\frac{x \cdot x_0}{a^2} + \frac{y \cdot y_0}{b^2} = 1$$

For hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the tangent at \((x_0, y_0)\) is:

$$\frac{x \cdot x_0}{a^2} - \frac{y \cdot y_0}{b^2} = 1$$

Example: Tangent Line Using Implicit Differentiation

Problem: Find the equation of the tangent line to the ellipse \(\frac{x^2}{9} + \frac{y^2}{4} = 1\) at the point \((3\cos\theta, 2\sin\theta)\) where \(\theta = \frac{\pi}{3}\).

Solution:

First, find the point: \(x_0 = 3\cos(\frac{\pi}{3}) = \frac{3}{2}\), \(y_0 = 2\sin(\frac{\pi}{3}) = \sqrt{3}\)
Use implicit differentiation on \(\frac{x^2}{9} + \frac{y^2}{4} = 1\):
$$\frac{2x}{9} + \frac{2y}{4}\frac{dy}{dx} = 0$$ $$\frac{dy}{dx} = -\frac{4x}{9y}$$
Evaluate at \((\frac{3}{2}, \sqrt{3})\):
$$m = -\frac{4 \cdot \frac{3}{2}}{9\sqrt{3}} = -\frac{6}{9\sqrt{3}} = -\frac{2}{3\sqrt{3}} = -\frac{2\sqrt{3}}{9}$$
Use point-slope form: \(y - \sqrt{3} = -\frac{2\sqrt{3}}{9}(x - \frac{3}{2})\)

Verification: Using the shortcut formula \(\frac{x \cdot x_0}{9} + \frac{y \cdot y_0}{4} = 1\): \(\frac{x \cdot \frac{3}{2}}{9} + \frac{y\sqrt{3}}{4} = 1\) gives the same line!

Reflective Properties of Conics

Parabola Reflective Property: Any ray parallel to the axis of a parabola reflects through the focus. This is why parabolic mirrors and satellite dishes work - they focus parallel incoming rays (light or radio waves) to a single point!
Ellipse Reflective Property: A ray from one focus reflects off the ellipse and passes through the other focus. This creates "whispering galleries" where sound from one focus point can be clearly heard at the other focus.
Hyperbola Reflective Property: A ray directed toward one focus reflects as if it came from the other focus. This property is used in telescope designs (Cassegrain reflector) and navigation systems (LORAN).

Parabola: Parallel Rays Focus

Parallel rays reflect through the focus.

Ellipse: Focus to Focus

Rays from one focus reflect to the other.

Historical Context
The reflective properties of conics have profound applications. Parabolic mirrors in telescopes focus parallel light rays to a point. Elliptical rooms create "whispering galleries" where sound from one focus is perfectly audible at the other. Hyperbolic mirrors are used in telescope designs (Cassegrain reflector) to create compact, high-performance optical systems. The Statuary Hall in the US Capitol is a famous elliptical whispering gallery.

Practice Problems: Intermediate Level

  1. Use the discriminant to identify: \(x^2 + 4xy + 4y^2 - 3x + 2y = 0\)
  2. Convert to standard form: \(x^2 - 4y^2 - 2x - 16y - 19 = 0\)
  3. Convert to standard form: \(9x^2 + 4y^2 + 36x - 8y + 4 = 0\)
  4. Find the tangent line to \(x^2 + y^2 = 25\) at \((3, 4)\).
  5. Find the tangent line to \(\frac{x^2}{9} - \frac{y^2}{16} = 1\) at \((5, \frac{16}{3})\).
ADVANCED LEVEL

Advanced Topics

Rotation of Axes

When a conic equation contains an \(xy\) term, the conic is rotated. We can eliminate the \(xy\) term by rotating the coordinate system through an angle \(\theta\).

Rotation Formulas: $$x = x'\cos\theta - y'\sin\theta$$ $$y = x'\sin\theta + y'\cos\theta$$

To eliminate the \(xy\) term in \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\), use:

$$\cot(2\theta) = \frac{A - C}{B}$$

Example 12: Eliminating the \(xy\) Term

Problem: Eliminate the \(xy\) term from \(xy = 4\) and identify the conic.

Solution:

Here \(A = 0\), \(B = 1\), \(C = 0\)
Find rotation angle:
$$\cot(2\theta) = \frac{0 - 0}{1} = 0$$ $$2\theta = 90°$$ $$\theta = 45°$$
Rotation formulas with \(\theta = 45°\):
$$x = \frac{x' - y'}{\sqrt{2}}, \quad y = \frac{x' + y'}{\sqrt{2}}$$
Substitute into \(xy = 4\):
$$\frac{x' - y'}{\sqrt{2}} \cdot \frac{x' + y'}{\sqrt{2}} = 4$$ $$\frac{(x')^2 - (y')^2}{2} = 4$$ $$(x')^2 - (y')^2 = 8$$ $$\frac{(x')^2}{8} - \frac{(y')^2}{8} = 1$$

This is a hyperbola rotated 45° with \(a = b = 2\sqrt{2}\).

Parametric Equations

Parametric equations express \(x\) and \(y\) as functions of a third variable (usually \(t\)).

Conic Parametric Equations Parameter Range
Circle
\(x^2 + y^2 = r^2\)
\(x = r\cos t\)
\(y = r\sin t\)
\(0 \leq t \lt 2\pi\)
Ellipse
\(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\)
\(x = a\cos t\)
\(y = b\sin t\)
\(0 \leq t \lt 2\pi\)
Parabola
\(y^2 = 4px\)
\(x = pt^2\)
\(y = 2pt\)
\(-\infty \lt t \lt \infty\)
Hyperbola
\(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\)
\(x = a\sec t\)
\(y = b\tan t\)
\(0 \leq t \lt 2\pi, t \neq \frac{\pi}{2}, \frac{3\pi}{2}\)

Example 13: Parametric to Cartesian

Problem: Convert the parametric equations \(x = 3\cos t\), \(y = 5\sin t\) to Cartesian form.

Solution:

From \(x = 3\cos t\), we get \(\cos t = \frac{x}{3}\)
From \(y = 5\sin t\), we get \(\sin t = \frac{y}{5}\)
Use the identity \(\cos^2 t + \sin^2 t = 1\):
$$\left(\frac{x}{3}\right)^2 + \left(\frac{y}{5}\right)^2 = 1$$ $$\frac{x^2}{9} + \frac{y^2}{25} = 1$$

This is an ellipse with \(a = 5\) (vertical) and \(b = 3\).

Polar Coordinate Representations

In polar coordinates with focus at the origin, all conics can be represented by a single unified equation.

Polar Form of Conics (Focus at Origin): $$r = \frac{ed}{1 \pm e\cos\theta} \quad \text{or} \quad r = \frac{ed}{1 \pm e\sin\theta}$$ where \(e\) is eccentricity and \(d\) is distance from focus to directrix.
Conic Type Eccentricity Polar Equation Example
Circle (special) \(e = 0\) \(r = a\) (constant)
Ellipse \(0 \lt e \lt 1\) \(r = \frac{3}{1 + 0.5\cos\theta}\)
Parabola \(e = 1\) \(r = \frac{2}{1 + \cos\theta}\)
Hyperbola \(e \gt 1\) \(r = \frac{6}{1 + 2\cos\theta}\)

Example 14: Identifying Conic from Polar Equation

Problem: Identify the conic \(r = \frac{12}{3 - 2\cos\theta}\) and find its eccentricity.

Solution:

Divide numerator and denominator by 3:
$$r = \frac{4}{1 - \frac{2}{3}\cos\theta}$$
Compare with standard form \(r = \frac{ed}{1 - e\cos\theta}\)
Eccentricity: \(e = \frac{2}{3}\)

Since \(0 \lt e \lt 1\), this is an ellipse with \(ed = 4\), so \(d = 6\).

Real-World Applications

Planetary Orbits (Kepler's Laws):
  • First Law: Planets move in elliptical orbits with the Sun at one focus
  • Second Law: A line from the Sun to a planet sweeps out equal areas in equal times
  • Third Law: The square of orbital period is proportional to the cube of the semi-major axis

Eccentricity determines orbit shape: Earth (\(e \approx 0.017\)) has nearly circular orbit, while Halley's Comet (\(e \approx 0.967\)) has highly elliptical orbit.

Parabolic Reflectors:
  • Satellite dishes: Collect parallel signals and focus them to receiver at focus
  • Headlights: Light source at focus produces parallel beam
  • Solar cookers: Focus sunlight to achieve high temperatures
  • Radio telescopes: Focus radio waves from space
Architectural Applications:
  • Elliptical arches: Distribute weight evenly (better than circular arches)
  • Whispering galleries: Elliptical domes allow sound to travel from one focus to another
  • Cooling towers: Hyperboloid structures provide strength with minimum material
Historical Context
The LORAN (Long Range Navigation) system used hyperbolic positioning. Ships and aircraft received signals from pairs of transmitter stations. The time difference between signals defined a hyperbola of possible locations. Using multiple station pairs, the intersection of hyperbolas gave the precise position. Though largely replaced by GPS, LORAN demonstrates elegant practical application of hyperbola mathematics.

Calculus with Conics

Arc Length of Ellipse:

For ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), the perimeter is approximately:

$$P \approx \pi(a + b)\left(1 + \frac{3h}{10 + \sqrt{4 - 3h}}\right)$$ where \(h = \frac{(a-b)^2}{(a+b)^2}\)

Note: There is no simple closed form for ellipse perimeter; this is Ramanujan's approximation.

Area Enclosed by Conic:
  • Circle: \(A = \pi r^2\)
  • Ellipse: \(A = \pi ab\)
  • Parabola (finite region): Use integration
Surface Area of Revolution:

Rotating a parabola \(y = x^2\) from \(x = 0\) to \(x = a\) about the \(y\)-axis creates a paraboloid:

$$S = \frac{\pi}{6}\left[(1 + 4a^2)^{3/2} - 1\right]$$

Degenerate Cases

When a plane intersects a double cone in special ways, we get degenerate conics:

Degenerate Case Geometric Description Algebraic Example
Point Plane through vertex only \(x^2 + y^2 = 0\) → point \((0,0)\)
Line Tangent plane to cone \((x - y)^2 = 0\) → line \(y = x\)
Two intersecting lines Plane through vertex and cone \(x^2 - y^2 = 0\) → lines \(y = \pm x\)
Empty set No real solutions \(x^2 + y^2 = -1\)

Degenerate cases are limiting forms of standard conics. For example, as the eccentricity of a hyperbola approaches infinity, it approaches its asymptotes (two intersecting lines).

Three-Dimensional Extensions: Quadric Surfaces

Conics extend naturally to three dimensions, creating quadric surfaces.

Surface Equation Description
Sphere \(x^2 + y^2 + z^2 = r^2\) 3D extension of circle
Ellipsoid \(\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1\) 3D extension of ellipse; all cross-sections are ellipses
Elliptic Paraboloid \(z = \frac{x^2}{a^2} + \frac{y^2}{b^2}\) Bowl-shaped; cross-sections are parabolas and ellipses
Hyperbolic Paraboloid \(z = \frac{x^2}{a^2} - \frac{y^2}{b^2}\) Saddle-shaped; "Pringles chip" surface
Hyperboloid (1 sheet) \(\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1\) Cooling tower shape; ruled surface
Hyperboloid (2 sheets) \(\frac{x^2}{a^2} - \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1\) Two separate surfaces facing away
Elliptic Cone \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = \frac{z^2}{c^2}\) Double cone; degenerate quadric
Applications of Quadric Surfaces:
  • Hyperboloid cooling towers: Structural strength with minimum material
  • Ellipsoid Earth model: More accurate than sphere for GPS
  • Paraboloid reflectors: 3D parabolic mirrors and antennas
  • Hyperbolic paraboloid roofs: Architecturally striking and structurally efficient

Practice Problems: Advanced Level

  1. Eliminate the \(xy\) term from \(5x^2 + 6xy + 5y^2 - 8 = 0\) and identify the conic.
  2. Write parametric equations for the hyperbola \(\frac{x^2}{16} - \frac{y^2}{9} = 1\).
  3. Identify the conic \(r = \frac{15}{5 - 3\sin\theta}\) and find its eccentricity.
  4. Find the area enclosed by the ellipse \(\frac{x^2}{25} + \frac{y^2}{16} = 1\).
  5. Identify the quadric surface: \(x^2 + y^2 - z^2 = 1\).

Comprehensive Comparison of All Conics

Property Circle Ellipse Parabola Hyperbola
Eccentricity \(e\) \(e = 0\) \(0 \lt e \lt 1\) \(e = 1\) \(e \gt 1\)
Standard Form \(x^2 + y^2 = r^2\) \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) \(x^2 = 4py\) \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\)
Number of Foci 1 (at center) 2 1 2
Focal Relationship — \(c^2 = a^2 - b^2\) — \(c^2 = a^2 + b^2\)
Asymptotes None None None \(y = \pm\frac{b}{a}x\)
Directrix At infinity 2 lines 1 line 2 lines
Polar Form \(r = a\) \(r = \frac{ed}{1 \pm e\cos\theta}\) \(r = \frac{d}{1 \pm \cos\theta}\) \(r = \frac{ed}{1 \pm e\cos\theta}\)
Parametric \(x = r\cos t\)
\(y = r\sin t\)
\(x = a\cos t\)
\(y = b\sin t\)
\(x = pt^2\)
\(y = 2pt\)
\(x = a\sec t\)
\(y = b\tan t\)
Key Application Wheels, pipes Planetary orbits Projectile motion Navigation (LORAN)

Conclusion: The Unity of Conic Sections

Throughout this tutorial, we've explored how conic sections—circles, ellipses, parabolas, and hyperbolas—form a unified family of curves, all arising from the intersection of a plane with a double cone.

The Power of Eccentricity: The single parameter \(e\) unifies all conics through the focus-directrix definition. As eccentricity increases from 0, we progress from circles to ellipses, then parabolas at \(e = 1\), and finally hyperbolas for \(e \gt 1\). This provides both geometric insight and computational power.

From ancient Greek mathematics to modern GPS systems, from Kepler's laws of planetary motion to satellite dish design, conic sections demonstrate the deep connection between pure mathematics and physical reality. The same equations that describe planetary orbits also govern the shape of a thrown ball, the design of telescope mirrors, and the architecture of cooling towers.

As you continue your mathematical journey, remember that these curves represent more than abstract equations—they are fundamental patterns in nature, technology, and art, revealing the mathematical structure underlying our universe.

Review Material for This Page

Definitions: The Coordinate Plane · Conic Sections

Formula quizzes: Coordinate Geometry Formulas · Conic Section Formulas

The synthetic groundwork this page builds on is in the Euclidean Geometry tutorial. Next in the study order is the Number Theory tutorial, which opens Tier 2.