Mathematics / Geometry 2D Coordinate Geometry & Conics 100% Free Open Access
Chapter 6 โ€ข Theory & Derivations

In-Depth Study of the Ellipse

Exhaustive treatment of the ellipse: focus-directrix definition e < 1 and canonical derivation x^2/a^2 + y^2/b^2 = 1; sum of focal distances SP + S'P = 2a and the gardener's construction; auxiliary circle x^2 + y^2 = a^2 and eccentric angle phi; area pi*a*b; tangents in point, slope, and parametric forms; director circle x^2 + y^2 = a^2 + b^2 and orthogonal tangent loci; normal equation a^2 x/x_1 - b^2 y/y_1 = a^2 - b^2; conjugate diameters and Apollonius' first (CP^2 + CD^2 = a^2 + b^2) and second (area = 4ab) theorems; optical and acoustic reflection properties; and the product of focal perpendiculars to any tangent p_1 p_2 = b^2.

ยง6.1 The Ellipse: Focus-Directrix Definition, Canonical Form & Metric Relations

1. The Focus-Directrix Definition ($e < 1$)

An ellipse is the planar locus of a point $P(x, y)$ that moves such that the ratio of its distance from a fixed focus $S(ae, 0)$ to its distance from a fixed directrix line $D: x = a/e$ is a constant eccentricity $e \in (0, 1)$: $$\frac{SP}{PM} = e \iff SP = e \cdot PM$$ Let $P(x, y)$ be any point on the curve. Then: $$SP^2 = e^2 PM^2 \implies (x - ae)^2 + y^2 = e^2 \left(x - \frac{a}{e}\right)^2 = (ex - a)^2$$ $$x^2 - 2aex + a^2 e^2 + y^2 = e^2 x^2 - 2aex + a^2$$ Canceling $-2aex$ and grouping terms: $$(1 - e^2)x^2 + y^2 = a^2(1 - e^2) \iff \frac{x^2}{a^2} + \frac{y^2}{a^2(1 - e^2)} = 1$$ Defining the minor semi-axis $b > 0$ by the fundamental relation: $$\mathbf{b^2 \equiv a^2(1 - e^2) \iff e = \sqrt{1 - \frac{b^2}{a^2}} < 1}$$ yields the universal canonical equation of the ellipse: $$\mathbf{\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \quad (a > b > 0)}$$

2. The Two Foci and The Constant Sum of Focal Radii

By symmetry about the $y$-axis, the ellipse possesses a second focus $S'(-ae, 0)$ and a second directrix $D': x = -a/e$. The focal distances to any point $P(x, y)$ on the ellipse are: $$SP = a - ex, \qquad S'P = a + ex$$ Adding the two distances: $$SP + S'P = (a - ex) + (a + ex) = \mathbf{2a = \text{Constant Everywhere!}}$$ The Gardener's / Focal Distance Theorem: An ellipse is the locus of all points whose sum of distances from two fixed foci $S$ and $S'$ is constant and equal to the major axis $2a$.

3. Canonical Geometric Elements

  • Center ($C$): Origin $(0, 0)$.
  • Major Axis: Segment $A'A$ along the $x$-axis, length $2a$.
  • Minor Axis: Segment $B'B$ along the $y$-axis, length $2b$.
  • Foci: $S(ae, 0)$ and $S'(-ae, 0)$, distance between foci $SS' = 2ae$.
  • Directrices: $x = \pm a/e$, distance between directrices $2a/e$.
  • Latus Rectum: Chord through focus perpendicular to major axis. Substituting $x = ae$: $$\frac{a^2 e^2}{a^2} + \frac{y^2}{b^2} = 1 \implies \frac{y^2}{b^2} = 1 - e^2 = \frac{b^2}{a^2} \implies y = \pm \frac{b^2}{a} \implies \mathbf{Length = \frac{2b^2}{a}}$$

ยง6.2 Auxiliary Circle, Eccentric Angle & Parametric Coordinates

1. The Auxiliary Circle

The circle described on the major axis $A'A$ of the ellipse as diameter is termed the auxiliary circle. Its equation is: $$x^2 + y^2 = a^2$$ Let $P(x, y)$ be any point on the ellipse. Draw a vertical ordinate through $P$ and extend it to meet the auxiliary circle at $Q(x, Y)$. Because $Q$ lies on the auxiliary circle, $x = a \cos \phi$, where $\phi$ is the angle $\angle OCQ$ measured from the major axis. Substituting $x = a \cos \phi$ into the ellipse equation: $$\frac{a^2 \cos^2 \phi}{a^2} + \frac{y^2}{b^2} = 1 \implies \cos^2 \phi + \frac{y^2}{b^2} = 1 \implies \frac{y^2}{b^2} = \sin^2 \phi \implies y = b \sin \phi$$ The angle $\phi \in [0, 2\pi)$ is termed the eccentric angle of point $P$.

2. Standard Parametric Form

The parametric coordinates of any point on the ellipse are: $$\mathbf{P(\phi) = (a \cos \phi, \; b \sin \phi)}$$ Notice the vertical scaling relation between the ellipse and its auxiliary circle: $$\frac{y_P}{Y_Q} = \frac{b \sin \phi}{a \sin \phi} = \frac{b}{a} = \text{Constant}$$ An ellipse is an auxiliary circle uniformly compressed vertically by the factor $b/a$! Consequently, the area of the ellipse is: $$\operatorname{Area}(\text{Ellipse}) = \frac{b}{a} \times \operatorname{Area}(\text{Auxiliary Circle}) = \frac{b}{a}(\pi a^2) = \mathbf{\pi a b}$$

ยง6.3 Tangents, Normals & The Director Circle

1. Equations of the Tangent

  1. Point Form: Tangent at $P(x_1, y_1)$ on the ellipse: $$\mathbf{\frac{x x_1}{a^2} + \frac{y y_1}{b^2} = 1}$$
  2. Parametric Form: Tangent at $P(\phi) = (a\cos\phi, b\sin\phi)$: $$\mathbf{\frac{x \cos \phi}{a} + \frac{y \sin \phi}{b} = 1}$$
  3. Slope Form: A line $y = mx + c$ is tangent to $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ if and only if $c^2 = a^2 m^2 + b^2$: $$\mathbf{y = m x \pm \sqrt{a^2 m^2 + b^2}}$$

2. The Director Circle of the Ellipse

Let $P(h, k)$ be the point of intersection of two mutually perpendicular tangents to the ellipse. The slope form of a tangent passing through $(h, k)$ is: $$k - mh = \pm \sqrt{a^2 m^2 + b^2} \iff (k - mh)^2 = a^2 m^2 + b^2$$ Expanding and grouping as a quadratic in the slope $m$: $$(h^2 - a^2)m^2 - 2hk m + (k^2 - b^2) = 0$$ If the two tangents are perpendicular, the product of their slopes must be $-1$: $$m_1 m_2 = \frac{k^2 - b^2}{h^2 - a^2} = -1 \iff k^2 - b^2 = -(h^2 - a^2) \iff h^2 + k^2 = a^2 + b^2$$ Replacing $(h, k)$ with current coordinates $(x, y)$: $$\mathbf{x^2 + y^2 = a^2 + b^2}$$ The Director Circle Theorem: The locus of the point of intersection of two mutually perpendicular tangents to an ellipse is a concentric circle of radius $R = \sqrt{a^2 + b^2}$, termed the director circle!

3. Equation of the Normal

The normal line at $P(x_1, y_1)$ is perpendicular to the tangent: $$\mathbf{\frac{a^2 x}{x_1} - \frac{b^2 y}{y_1} = a^2 - b^2}$$ In parametric coordinates $P(\phi)$: $$\mathbf{a x \sec \phi - b y \csc \phi = a^2 - b^2}$$

ยง6.4 Conjugate Diameters & Apollonius' Theorems

1. Definition of Conjugate Diameters

A diameter of an ellipse is a chord passing through its center $C(0, 0)$. Two diameters $y = m_1 x$ and $y = m_2 x$ are said to be conjugate diameters if each bisects all chords parallel to the other. The algebraic condition for conjugacy is: $$\mathbf{m_1 m_2 = -\frac{b^2}{a^2}}$$ In parametric terms, if $CP$ is a semi-diameter with endpoint $P(\phi) = (a\cos\phi, b\sin\phi)$, its conjugate semi-diameter $CD$ has endpoint $D$ whose eccentric angle differs by $\pi/2$: $$\phi_D = \phi + \frac{\pi}{2} \implies \mathbf{D = (-a \sin \phi, \; b \cos \phi)}$$

2. Apollonius' First Theorem (Sum of Squared Semi-Diameters)

The sum of the squares of any two conjugate semi-diameters is constant and equal to the sum of the squares of the semi-axes: $$CP^2 = a^2 \cos^2 \phi + b^2 \sin^2 \phi$$ $$CD^2 = a^2 \sin^2 \phi + b^2 \cos^2 \phi$$ Adding the two equations: $$CP^2 + CD^2 = a^2(\cos^2 \phi + \sin^2 \phi) + b^2(\sin^2 \phi + \cos^2 \phi) = \mathbf{a^2 + b^2 = \text{Constant!}}$$

3. Apollonius' Second Theorem (Area of Circumscribing Parallelogram)

The area of the parallelogram formed by the tangents drawn at the extremities of any pair of conjugate diameters is constant and equal to the area of the rectangle formed by the principal axes: $$\mathcal{A} = 4 \left| x_P y_D - x_D y_P \right| = 4 |(a\cos\phi)(b\cos\phi) - (-a\sin\phi)(b\sin\phi)| = 4 a b(\cos^2 \phi + \sin^2 \phi) = \mathbf{4 a b}$$

ยง6.5 Optical Reflection Property & Product of Focal Perpendiculars

1. The Optical/Acoustic Reflection Property

Let $P(x_1, y_1)$ be any point on the ellipse with foci $S(ae, 0)$ and $S'(-ae, 0)$. Let the normal at $P$ meet the major axis at $G$. By the properties of the normal: $$CG = e^2 x_1 \implies SG = ae - e^2 x_1 = e(a - ex_1) = e \cdot SP, \quad S'G = ae + e^2 x_1 = e(a + ex_1) = e \cdot S'P$$ Therefore: $$\frac{SG}{S'G} = \frac{SP}{S'P}$$ By the angle bisector theorem, the normal $PG$ is the internal bisector of the focal angle $\angle SPS'$! Consequently, the tangent at $P$ is the external bisector of $\angle SPS'$. $$\mathbf{\angle S P T = \angle S' P T}$$ Physical Consequence: Any light ray or acoustic wave emitted from one focus $S$ reflects off the elliptical boundary directly to the other focus $S'$! This is the physical mechanism of whispering galleries (such as St. Paul's Cathedral in London and the National Statuary Hall in Washington, D.C.).

2. Product of Perpendiculars from Foci onto Any Tangent

Let $p_1$ and $p_2$ be the lengths of perpendiculars dropped from the two foci $S(ae, 0)$ and $S'(-ae, 0)$ onto any tangent line $y - mx - \sqrt{a^2 m^2 + b^2} = 0$: $$p_1 = \frac{|-mae - \sqrt{a^2 m^2 + b^2}|}{\sqrt{1 + m^2}}, \qquad p_2 = \frac{|mae - \sqrt{a^2 m^2 + b^2}|}{\sqrt{1 + m^2}}$$ Multiplying the two perpendiculars: $$p_1 p_2 = \frac{|(a^2 m^2 + b^2) - m^2 a^2 e^2|}{1 + m^2} = \frac{|a^2 m^2(1 - e^2) + b^2|}{1 + m^2} = \frac{|a^2 m^2 (b^2/a^2) + b^2|}{1 + m^2} = \frac{b^2(m^2 + 1)}{1 + m^2} = \mathbf{b^2}$$ Theorem: The product of the perpendiculars from the foci onto any tangent to an ellipse is constant and equal to the square of the semi-minor axis $b^2$!

TIERED UNIVERSITY HONORS PROBLEMS

Step-by-Step Solved Examination Problems

Comprehensive analytical derivations, multi-tier solutions (Foundational, Intermediate Exam, and Honors/Proof Challenge) with complete line-by-line verification.

Tier 1: Foundational Example 6.1: Standard Ellipse Geometric Elements and Parametric Tangent

For the ellipse $9x^2 + 16y^2 = 144$: (a) Find the lengths of the major and minor axes, eccentricity $e$, and coordinates of the foci and directrices. (b) Find the equation of the tangent line at the point where the eccentric angle is $\phi = \pi/4$.

Step 1: Reduce to Canonical Form
$$\frac{9x^2}{144} + \frac{16y^2}{144} = 1 \implies \frac{x^2}{16} + \frac{y^2}{9} = 1 \implies a^2 = 16, \; b^2 = 9 \implies a = 4, \; b = 3$$

Major axis length is $2a = 8$; minor axis length is $2b = 6$.

Step 2: Compute Eccentricity, Foci and Directrices
$$e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{9}{16}} = \frac{\sqrt{7}}{4} \\ \text{Foci: } (\pm ae, 0) = \left(\pm 4\cdot\frac{\sqrt{7}}{4}, 0\right) = (\pm\sqrt{7}, 0) \\ \text{Directrices: } x = \pm\frac{a}{e} = \pm\frac{4}{\sqrt{7}/4} = \pm\frac{16}{\sqrt{7}}$$

Focal distance is $ae = \sqrt{7}$ and directrix distance is $a/e = 16/\sqrt{7}$.

Step 3: Tangent Equation at $\phi = \pi/4$
$$\frac{x\cos\phi}{a} + \frac{y\sin\phi}{b} = 1 \implies \frac{x\cos(\pi/4)}{4} + \frac{y\sin(\pi/4)}{3} = 1 \implies \frac{x}{4\sqrt{2}} + \frac{y}{3\sqrt{2}} = 1 \\ 3x + 4y = 12\sqrt{2}$$

Multiplying by $12\sqrt{2}$ yields the linear tangent equation $3x + 4y - 12\sqrt{2} = 0$.

Final Answer & Physical Insight

a = 4, \; b = 3; \quad e = \frac{\sqrt{7}}{4}; \quad \text{Foci: } (\pm\sqrt{7}, 0); \quad \text{Directrices: } x = \pm\frac{16}{\sqrt{7}}; \quad \text{Tangent: } 3x + 4y = 12\sqrt{2}

Tier 2: Intermediate Exam Example 6.2: Director Circle and Mutual Perpendicular Tangents

Given the ellipse $\frac{x^2}{25} + \frac{y^2}{9} = 1$: (a) Write the equation of its director circle. (b) Find the equations of the tangents drawn from the point $P(0, \sqrt{34})$. (c) Verify that the two tangents are mutually perpendicular.

Step 1: Equation of Director Circle
$$x^2 + y^2 = a^2 + b^2 = 25 + 9 = 34 \implies x^2 + y^2 = 34$$

Notice that the point $P(0, \sqrt{34})$ satisfies $0^2 + (\sqrt{34})^2 = 34$, so $P$ lies strictly on the director circle!

Step 2: Slope Form of Tangents through $P(0, \sqrt{34})$
$$y = mx \pm \sqrt{a^2 m^2 + b^2} = mx \pm \sqrt{25m^2 + 9} \\ \sqrt{34} = m(0) \pm \sqrt{25m^2 + 9} \implies 34 = 25m^2 + 9 \implies 25m^2 = 25 \implies m^2 = 1 \implies m = \pm 1$$

The slopes of the two tangents are $m_1 = 1$ and $m_2 = -1$.

Step 3: Tangent Equations and Orthogonality
$$\text{Tangent 1 } (m = 1): \quad y = x + \sqrt{34} \implies x - y + \sqrt{34} = 0 \\ \text{Tangent 2 } (m = -1): \quad y = -x + \sqrt{34} \implies x + y - \sqrt{34} = 0 \\ m_1 \cdot m_2 = (1)(-1) = -1 \implies \text{Strictly Perpendicular!}$$

The product of the slopes is $-1$, verifying the Director Circle Theorem.

Final Answer & Physical Insight

\text{Director Circle: } x^2 + y^2 = 34; \quad \text{Tangents: } y = x + \sqrt{34} \text{ and } y = -x + \sqrt{34}; \quad m_1 m_2 = -1 \implies \text{Orthogonal}

Tier 3: Honors / Proof Challenge Example 6.3: Rigorous Proof of Apollonius' Conjugate Diameter Theorems

Let $CP$ and $CD$ be a pair of conjugate semi-diameters of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Prove analytically: (a) $CP^2 + CD^2 = a^2 + b^2$ (Apollonius' First Theorem). (b) The area of the parallelogram formed by the tangents at the extremities of the conjugate diameters is constant and equal to $4ab$ (Apollonius' Second Theorem).

Step 1: Express Endpoints in Parametric Form
$$\text{Let } P = (a\cos\phi, b\sin\phi). \quad \text{Then } D \text{ has eccentric angle } \phi + \pi/2: \\ D = (a\cos(\phi + \pi/2), b\sin(\phi + \pi/2)) = (-a\sin\phi, b\cos\phi)$$

This establishes the exact coordinates of conjugate extremities.

Step 2: Proof of Apollonius' First Theorem
$$CP^2 = (a\cos\phi)^2 + (b\sin\phi)^2 = a^2\cos^2\phi + b^2\sin^2\phi \\ CD^2 = (-a\sin\phi)^2 + (b\cos\phi)^2 = a^2\sin^2\phi + b^2\cos^2\phi \\ CP^2 + CD^2 = a^2(\cos^2\phi + \sin^2\phi) + b^2(\sin^2\phi + \cos^2\phi) = a^2(1) + b^2(1) = a^2 + b^2$$

Since $\phi$ cancels completely, $CP^2 + CD^2 = a^2 + b^2$ holds for every conjugate pair.

Step 3: Proof of Apollonius' Second Theorem
$$\text{Area of } \triangle CPD = \frac{1}{2}|x_P y_D - x_D y_P| = \frac{1}{2}|(a\cos\phi)(b\cos\phi) - (-a\sin\phi)(b\sin\phi)| \\ = \frac{1}{2}|ab\cos^2\phi + ab\sin^2\phi| = \frac{1}{2}ab(\cos^2\phi + \sin^2\phi) = \frac{1}{2}ab \\ \text{Total parallelogram area} = 8 \times \operatorname{Area}(\triangle CPD) \text{ (or } 4 \times 2\triangle) = 4ab$$

The circumscribing parallelogram has constant area $4ab$ everywhere.

Final Answer & Physical Insight

CP^2 + CD^2 = a^2 + b^2 \text{ (Apollonius' 1st)}; \quad \text{Parallelogram Area} = 4ab \text{ (Apollonius' 2nd)}