Conic Sections

CONIC SECTIONS

Introduction and Basic Concepts

**Conic sections** are curves obtained by intersecting a right circular cone with a plane. The term "conic sections" comes from the fact that these curves can be generated by cutting a double-napped right circular cone at different angles.

The key components of a cone are:

**Vertex (V)**: The fixed point where two nappes meet

**Axis (l)**: The fixed vertical line passing through the vertex

**Generator (m)**: The rotating line that maintains a constant angle α with the axis

**Nappe**: One half of the double cone, separated by the vertex

The curves obtained from conic sections have extensive real-world applications: planetary orbits follow elliptical paths, parabolas are used in the design of reflectors for telescopes and automobile headlights, hyperbolas appear in navigation systems, and circles form the basis of many engineering designs.

Sections of a Cone

When a plane intersects a right circular cone, the type of curve formed depends on the angle β that the plane makes with the vertical axis of the cone, relative to the semi-vertical angle α of the cone.

Non-degenerate conic sections (plane cuts through nappe, not at vertex):

**Circle**: Formed when **β = 90°** (plane perpendicular to axis)

The plane cuts entirely across one nappe

All points on the intersection are equidistant from a center point

**Ellipse**: Formed when **α < β < 90°** (plane oblique to axis)

The plane cuts entirely across one nappe at an angle

Creates an oval-shaped closed curve

**Parabola**: Formed when **β = α** (plane parallel to generator)

The plane is parallel to exactly one generator of the cone

Creates an open curve extending infinitely

**Hyperbola**: Formed when **0 ≤ β < α** (plane cuts both nappes)

The plane intersects both nappes of the cone

Creates two separate open curves (branches)

Degenerate conic sections (plane cuts at vertex):

**Point**: Formed when **α < β ≤ 90°**

Degenerate case of a circle or ellipse

**Single straight line**: Formed when **β = α**

Degenerate case of a parabola

The plane contains a generator of the cone

**Pair of intersecting lines**: Formed when **0 ≤ β < α**

Degenerate case of a hyperbola

Two lines intersect at the vertex

Circle

Definition and Standard Equation

**Definition**: A circle is the set of all points in a plane that are equidistant from a fixed point (center) in the plane. The constant distance is called the radius.

**Standard equation of circle with center (h, k) and radius r**:

$$(x - h)^2 + (y - k)^2 = r^2$$

**Derivation**: Let C(h, k) be the center and P(x, y) be any point on the circle. By the distance formula and the definition of circle:

$$\sqrt{(x-h)^2 + (y-k)^2} = r$$

Squaring both sides gives the standard form.

**Special case**: When center is at origin (0, 0):

$$x^2 + y^2 = r^2$$

Converting General Form to Standard Form

Given a circle equation in general form like $x^2 + y^2 + 2gx + 2fy + c = 0$:

1. Rearrange: $(x^2 + 2gx) + (y^2 + 2fy) = -c$

2. Complete the square for both variables:

$(x^2 + 2gx + g^2) + (y^2 + 2fy + f^2) = g^2 + f^2 - c$

3. Factor: $(x + g)^2 + (y + f)^2 = g^2 + f^2 - c$

From this form:

Center = $(-g, -f)$

Radius = $\sqrt{g^2 + f^2 - c}$

Worked Examples

**Example 1**: Find the equation of circle with center (−3, 2) and radius 4.

Using $(x - h)^2 + (y - k)^2 = r^2$:

$(x - (-3))^2 + (y - 2)^2 = 4^2$

$(x + 3)^2 + (y - 2)^2 = 16$

**Example 2**: Find center and radius of $x^2 + y^2 + 8x + 10y - 8 = 0$.

Rearranging: $(x^2 + 8x) + (y^2 + 10y) = 8$

Completing squares:

$(x^2 + 8x + 16) + (y^2 + 10y + 25) = 8 + 16 + 25$

$(x + 4)^2 + (y + 5)^2 = 49 = 7^2$

Therefore: Center = (−4, −5), Radius = 7

**Example 3**: Find equation of circle passing through (2, −2) and (3, 4) with center on line x + y = 2.

Let center be (h, k). Using the condition that distances from center to both points are equal:

$(2 - h)^2 + (-2 - k)^2 = (3 - h)^2 + (4 - k)^2$

Expanding and simplifying:

$4 - 4h + h^2 + 4 + 4k + k^2 = 9 - 6h + h^2 + 16 - 8k + k^2$

$8 - 4h + 4k = 25 - 6h - 8k$

$2h + 12k = 17$ ... (1)

Also, center lies on x + y = 2:

$h + k = 2$ ... (2)

From (2): $h = 2 - k$

Substitute in (1): $2(2 - k) + 12k = 17$

$4 - 2k + 12k = 17$

$10k = 13$, so $k = 1.3$, $h = 0.7$

Radius: $r^2 = (2 - 0.7)^2 + (-2 - 1.3)^2 = 1.69 + 10.89 = 12.58$

Equation: $(x - 0.7)^2 + (y - 1.3)^2 = 12.58$

Parabola

Definition and Key Terms

**Definition**: A parabola is the set of all points in a plane that are equidistant from a fixed line (directrix) and a fixed point (focus) not on the line.

**Key components**:

**Focus**: The fixed point F

**Directrix**: The fixed line

**Axis**: The line through the focus perpendicular to the directrix

**Vertex**: The point where the parabola intersects its axis (midpoint between focus and directrix)

**Important property**: For any point P on the parabola: **Distance from P to focus = Distance from P to directrix**

Standard Equations of Parabola

There are four standard forms based on the position of focus and direction of opening:

**Form 1**: Focus at (a, 0), Directrix x = −a, Opens RIGHT

$$y^2 = 4ax \quad (a > 0)$$

**Form 2**: Focus at (−a, 0), Directrix x = a, Opens LEFT

$$y^2 = -4ax \quad (a > 0)$$

**Form 3**: Focus at (0, a), Directrix y = −a, Opens UPWARD

$$x^2 = 4ay \quad (a > 0)$$

**Form 4**: Focus at (0, −a), Directrix y = a, Opens DOWNWARD

$$x^2 = -4ay \quad (a > 0)$$

Derivation of Standard Form

For parabola $y^2 = 4ax$ with focus F(a, 0) and directrix x = −a:

Let P(x, y) be any point on the parabola. Let B(−a, y) be the foot of perpendicular from P to the directrix.

By definition: $PF = PB$

$$\sqrt{(x-a)^2 + y^2} = \sqrt{(x+a)^2 + 0^2}$$

Squaring both sides:

$(x - a)^2 + y^2 = (x + a)^2$

$x^2 - 2ax + a^2 + y^2 = x^2 + 2ax + a^2$

$y^2 = 4ax$

Properties of Parabola

1. **Symmetry**: Parabola is symmetric about its axis

If equation contains $y^2$, axis is along x-axis

If equation contains $x^2$, axis is along y-axis

2. **Direction of opening**:

When axis is x-axis: Opens RIGHT if coefficient of x is positive; opens LEFT if negative

When axis is y-axis: Opens UPWARD if coefficient of y is positive; opens DOWNWARD if negative

3. **Domain and Range**:

For $y^2 = 4ax$: x ≥ 0, all values of y

For $x^2 = 4ay$: y ≥ 0, all values of x

Latus Rectum

**Definition**: The latus rectum is a chord perpendicular to the axis passing through the focus, with both endpoints on the parabola.

**Length of latus rectum** for all four standard forms: **L = 4a**

Derivation for $y^2 = 4ax$: The endpoints of latus rectum are where x = a. Substituting into $y^2 = 4ax$:

$y^2 = 4a(a) = 4a^2$

$y = ±2a$

Length = $2a - (-2a) = 4a$

Worked Examples

**Example 1**: Find focus, directrix, axis, and latus rectum of $y^2 = 8x$.

Comparing with $y^2 = 4ax$: $4a = 8$, so $a = 2$

Focus: (2, 0)

Directrix: x = −2

Axis: x-axis (y = 0)

Latus rectum length: 4a = 8

Parabola opens: RIGHT (coefficient of x is positive)

**Example 2**: Find equation of parabola with vertex at (0, 0) and focus at (0, 2).

Since focus is on y-axis at (0, 2), equation is of form $x^2 = 4ay$ with a = 2.

$x^2 = 4(2)y$

$x^2 = 8y$

**Example 3**: Find equation of parabola symmetric about y-axis, passing through (2, −3).

Symmetric about y-axis with vertex at origin: form is $x^2 = ±4ay$

Since (2, −3) is in 4th quadrant and parabola passes through it, parabola opens DOWNWARD.

Form: $x^2 = -4ay$

Substituting (2, −3):

$4 = -4a(-3) = 12a$

$a = 1/3$

Equation: $x^2 = -4(1/3)y$

$x^2 = -(4/3)y$

$3x^2 = -4y$

Ellipse

Definition and Key Terms

**Definition**: An ellipse is the set of all points in a plane such that the **sum of distances from any point to two fixed points (foci) is constant**.

Let this constant sum = 2a, and distance between foci = 2c, then the constant is always > 2c.

**Key components**:

**Foci**: Two fixed points F₁ and F₂

**Center**: Midpoint of the line segment joining the foci

**Major axis**: Line segment through foci; length = 2a (longest diameter)

**Minor axis**: Line segment perpendicular to major axis through center; length = 2b

**Vertices**: Endpoints of major axis

**Semi-major axis**: Length a (half of major axis)

**Semi-minor axis**: Length b (half of minor axis)

Relationship Between a, b, and c

For any ellipse:

$$a^2 = b^2 + c^2$$

or equivalently:

$$c = \sqrt{a^2 - b^2}$$

**Derivation**: Consider point P at end of major axis:

Sum of distances = F₁P + F₂P = (c + a) + (a − c) = 2a

Consider point Q at end of minor axis:

Sum of distances = F₁Q + F₂Q = $2\sqrt{b^2 + c^2}$

Since both points on ellipse:

$2\sqrt{b^2 + c^2} = 2a$

$\sqrt{b^2 + c^2} = a$

$b^2 + c^2 = a^2$ ✓

Eccentricity

**Definition**: Eccentricity (e) is the ratio of distance from center to focus and distance from center to vertex:

$$e = \frac{c}{a}$$

**Properties of eccentricity**:

For ellipse: **0 < e < 1**

When e → 0: ellipse approaches a circle

When e → 1: ellipse becomes increasingly elongated

Focus distance from center: ae

Standard Equations of Ellipse

**Form 1**: Major axis along x-axis (a > b)

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

Where:

Center: (0, 0)

Foci: (±c, 0) where $c = \sqrt{a^2 - b^2}$

Vertices: (±a, 0)

Co-vertices: (0, ±b)

**Form 2**: Major axis along y-axis (a > b)

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

Where:

Center: (0, 0)

Foci: (0, ±c) where $c = \sqrt{a^2 - b^2}$

Vertices: (0, ±a)

Co-vertices: (±b, 0)

Derivation of Standard Form

For ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with foci at (−c, 0) and (c, 0):

Let P(x, y) be any point on ellipse. By definition:

$$PF_1 + PF_2 = 2a$$

$$\sqrt{(x+c)^2 + y^2} + \sqrt{(x-c)^2 + y^2} = 2a$$

Let $\sqrt{(x+c)^2 + y^2} = r_1$ and $\sqrt{(x-c)^2 + y^2} = r_2$

From $r_1 + r_2 = 2a$:

$r_1 = 2a - r_2$

Squaring: $(x+c)^2 + y^2 = 4a^2 - 4ar_2 + (x-c)^2 + y^2$

Simplifying: $4cx = 4a^2 - 4ar_2$

$r_2 = a - \frac{cx}{a}$

Similarly: $r_1 = a + \frac{cx}{a}$

Using $(x-c)^2 + y^2 = r_2^2$:

$(x-c)^2 + y^2 = (a - \frac{cx}{a})^2$

Expanding and simplifying:

$x^2 - 2cx + c^2 + y^2 = a^2 - 2cx + \frac{c^2x^2}{a^2}$

$x^2 + y^2 - \frac{c^2x^2}{a^2} = a^2 - c^2$

$x^2(1 - \frac{c^2}{a^2}) + y^2 = a^2 - c^2$

$x^2 \cdot \frac{a^2 - c^2}{a^2} + y^2 = a^2 - c^2$

Since $b^2 = a^2 - c^2$:

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ ✓

Properties of Ellipse

1. **Symmetry**: Ellipse is symmetric about both coordinate axes

If (x, y) on ellipse, then (−x, y), (x, −y), and (−x, −y) are also on ellipse

2. **Extent**: Ellipse lies within the rectangle bounded by x = ±a and y = ±b

3. **Identifying major axis**:

In $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$: Major axis is along x-axis if a > b

In $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$: Major axis is along y-axis (a is always semi-major)

The larger denominator indicates the direction of major axis

4. **Foci always lie on major axis**

Latus Rectum

**Definition**: Latus rectum is a chord perpendicular to major axis through a focus with both endpoints on the ellipse.

**Length of latus rectum** for all ellipses:

$$L = \frac{2b^2}{a}$$

**Derivation** for $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with focus at (c, 0):

Substitute x = c into ellipse equation:

$\frac{c^2}{a^2} + \frac{y^2}{b^2} = 1$

$\frac{y^2}{b^2} = 1 - \frac{c^2}{a^2} = \frac{a^2 - c^2}{a^2} = \frac{b^2}{a^2}$

$y^2 = \frac{b^4}{a^2}$

$y = ±\frac{b^2}{a}$

Length of latus rectum = $\frac{2b^2}{a}$

Relationship Between Parameters

For an ellipse with semi-major axis a, semi-minor axis b:

$c = \sqrt{a^2 - b^2}$ (distance of focus from center)

$e = \frac{c}{a} = \frac{\sqrt{a^2-b^2}}{a}$ (eccentricity)

Also: $e = \sqrt{1 - \frac{b^2}{a^2}}$

From e and a, we get: $b = a\sqrt{1 - e^2}$

Worked Examples

**Example 1**: For ellipse $\frac{x^2}{25} + \frac{y^2}{16} = 1$, find foci, vertices, and eccentricity.

Comparing with standard form:

$a^2 = 25$, so $a = 5$ (major axis along x-axis)

$b^2 = 16$, so $b = 4$

$c = \sqrt{a^2 - b^2} = \sqrt{25 - 16} = \sqrt{9} = 3$

Results:

Vertices: (±5, 0)

Co-vertices: (0, ±4)

Foci: (±3, 0)

Eccentricity: $e = \frac{c}{a} = \frac{3}{5} = 0.6$

Latus rectum: $L = \frac{2b^2}{a} = \frac{2(16)}{5} = \frac{32}{5} = 6.4$

**Example 2**: Find equation of ellipse with foci at (0, ±4) and semi-major axis = 5.

Since foci are on y-axis, major axis is along y-axis.

a = 5 (semi-major)

c = 4 (focus distance)

$b^2 = a^2 - c^2 = 25 - 16 = 9$, so b = 3

Equation: $\frac{x^2}{9} + \frac{y^2}{25} = 1$

**Example 3**: Find equation of ellipse with vertices at (±6, 0) and eccentricity = 2/3.

a = 6 (semi-major from vertices)

$e = \frac{2}{3}$, so $c = ae = 6 \times \frac{2}{3} = 4$

$b^2 = a^2 - c^2 = 36 - 16 = 20$

Equation: $\frac{x^2}{36} + \frac{y^2}{20} = 1$

Hyperbola

Definition and Key Terms

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

Let this constant = 2a, and distance between foci = 2c, then the constant is always < 2c.

**Key components**:

**Foci**: Two fixed points F₁ and F₂ on the transverse axis

**Center**: Midpoint of the line segment joining the foci

**Transverse axis**: Line segment through foci; length = 2a

**Conjugate axis**: Line segment perpendicular to transverse axis through center; length = 2b

**Vertices**: Endpoints of transverse axis; distance from center = a

**Semi-transverse axis**: Length a

**Semi-conjugate axis**: Length b

Relationship Between a, b, and c

For any hyperbola:

$$c^2 = a^2 + b^2$$

or equivalently:

$$c = \sqrt{a^2 + b^2}$$

**Derivation**: Consider point P at vertex (closest to center on transverse axis):

Difference of distances = |PF₁ − PF₂| = |(−a − c) − (a − c)| = 2a (taking absolute value correctly)

More precisely, for point at (a, 0):

Distance to F₁(−c, 0): $(a + c)$

Distance to F₂(c, 0): $(c − a)$

Difference: $(a + c) − (c − a) = 2a$ ✓

For any point on hyperbola, the sum of squares relation gives: $c^2 = a^2 + b^2$

Eccentricity

**Definition**: For hyperbola:

$$e = \frac{c}{a}$$

**Properties of eccentricity**:

For hyperbola: **e > 1**

The larger the eccentricity, the more "open" the branches

Minimum value of e: approaches 1 (as b → 0, hyperbola becomes narrower)

Focus distance from center: ae

Standard Equations of Hyperbola

**Form 1**: Transverse axis along x-axis

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

Where:

Center: (0, 0)

Vertices: (±a, 0)

Foci: (±c, 0) where $c = \sqrt{a^2 + b^2}$

Transverse axis along: x-axis

Conjugate axis along: y-axis

**Form 2**: Transverse axis along y-axis

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

Where:

Center: (0, 0)

Vertices: (0, ±a)

Foci: (0, ±c) where $c = \sqrt{a^2 + b^2}$

Transverse axis along: y-axis

Conjugate axis along: x-axis

Derivation of Standard Form

For hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ with foci at (−c, 0) and (c, 0):

Let P(x, y) be any point on hyperbola. By definition:

$$|PF_1 - PF_2| = 2a$$

For right branch (x > 0): $PF_1 - PF_2 = 2a$

$$\sqrt{(x+c)^2 + y^2} - \sqrt{(x-c)^2 + y^2} = 2a$$

Let $r_1 = \sqrt{(x+c)^2 + y^2}$ and $r_2 = \sqrt{(x-c)^2 + y^2}$

From $r_1 - r_2 = 2a$:

$r_1 = 2a + r_2$

Squaring: $(x+c)^2 + y^2 = 4a^2 + 4ar_2 + (x-c)^2 + y^2$

Simplifying: $4cx = 4a^2 + 4ar_2$

$r_2 = \frac{cx - a^2}{a}$

From $(x-c)^2 + y^2 = r_2^2$:

$(x-c)^2 + y^2 = (\frac{cx - a^2}{a})^2$

Expanding and simplifying:

$x^2 - 2cx + c^2 + y^2 = \frac{(cx - a^2)^2}{a^2}$

$a^2(x^2 - 2cx + c^2 + y^2) = (cx - a^2)^2$

$a^2x^2 - 2a^2cx + a^2c^2 + a^2y^2 = c^2x^2 - 2a^2cx + a^4$

$a^2x^2 + a^2c^2 + a^2y^2 = c^2x^2 + a^4$

$a^2x^2 - c^2x^2 + a^2y^2 = a^4 - a^2c^2$

$x^2(a^2 - c^2) + a^2y^2 = a^2(a^2 - c^2)$

Since $c^2 = a^2 + b^2$, we have $a^2 - c^2 = -b^2$:

$-b^2x^2 + a^2y^2 = -a^2b^2$

$b^2x^2 - a^2y^2 = a^2b^2$

$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ ✓

Properties of Hyperbola

1. **Symmetry**: Hyperbola is symmetric about both coordinate axes and about the center

If (x, y) on hyperbola, then (−x, y), (x, −y), and (−x, −y) are also on hyperbola

2. **Extent**:

For $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$: x ≤ −a or x ≥ a (two separate branches)

For $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$: y ≤ −a or y ≥ a (two separate branches)

3. **Asymptotes**:

For $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$: asymptotes are $y = ±\frac{b}{a}x$

For $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$: asymptotes are $y = ±\frac{a}{b}x$

The hyperbola approaches these lines but never reaches them as |x| or |y| → ∞

4. **Identifying transverse axis**:

In $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$: Transverse axis is along x-axis

In $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$: Transverse axis is along y-axis

The positive term indicates the direction

5. **Vertices and Foci**:

Vertices are on the transverse axis at distance a from center

Foci are on the transverse axis at distance c = √(a² + b²) from center

Latus Rectum

**Definition**: Latus rectum is a chord perpendicular to transverse axis through a focus with both endpoints on the hyperbola.

**Length of latus rectum** for all hyperbolas:

$$L = \frac{2b^2}{a}$$

**Derivation** for $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ with focus at (c, 0):

Substitute x = c into hyperbola equation:

$\frac{c^2}{a^2} - \frac{y^2}{b^2} = 1$

$\frac{y^2}{b^2} = \frac{c^2}{a^2} - 1 = \frac{c^2 - a^2}{a^2} = \frac{b^2}{a^2}$

$y^2 = \frac{b^4}{a^2}$

$y = ±\frac{b^2}{a}$

Length of latus rectum = $\frac{2b^2}{a}$ (same as ellipse!)

Worked Examples

**Example 1**: For hyperbola $\frac{x^2}{16} - \frac{y^2}{9} = 1$, find vertices, foci, eccentricity, and asymptotes.

Comparing with standard form:

$a^2 = 16$, so $a = 4$ (transverse axis along x-axis)

$b^2 = 9$, so $b = 3$

$c = \sqrt{a^2 + b^2} = \sqrt{16 + 9} = \sqrt{25} = 5$

Results:

Vertices: (±4, 0)

Foci: (±5, 0)

Eccentricity: $e = \frac{c}{a} = \frac{5}{4} = 1.25$

Asymptotes: $y = ±\frac{3}{4}x$

Latus rectum: $L = \frac{2b^2}{a} = \frac{2(9)}{4} = \frac{9}{2} = 4.5$

**Example 2**: Find equation of hyperbola with vertices at (0, ±5) and foci at (0, ±7).

Since vertices and foci are on y-axis, transverse axis is along y-axis.

a = 5 (semi-transverse)

c = 7 (focus distance)

$b^2 = c^2 - a