Mathematics / Geometry 2D Coordinate Geometry & Conics 100% Free Open Access
Chapter 7 β€’ Theory & Derivations

In-Depth Study of the Hyperbola & Rectangular Hyperbola

Exhaustive treatment of the hyperbola: focus-directrix definition e > 1 and canonical derivation x^2/a^2 - y^2/b^2 = 1; focal distance difference |S'P - SP| = 2a; asymptotes y = +/- (b/a)x; conjugate hyperbola y^2/b^2 - x^2/a^2 = 1 and the eccentricity relation 1/e_1^2 + 1/e_2^2 = 1; tangents in point, slope, and parametric forms; director circle x^2 + y^2 = a^2 - b^2; rectangular (equilateral) hyperbola x^2 - y^2 = a^2 with e = sqrt(2); rotation to asymptotic canonical form xy = c^2; and the constant triangle area 2c^2 and midpoint bisection properties.

Β§7.1 The Hyperbola: Canonical Equation, Eccentricity & Metric Relations

1. The Focus-Directrix Locus Definition ($e > 1$)

A hyperbola 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 $e > 1$: $$\frac{SP}{PM} = e \iff SP = e \cdot PM$$ Using the Euclidean distance formula: $$(x - ae)^2 + y^2 = e^2 \left(x - \frac{a}{e}\right)^2 = (ex - a)^2$$ Expanding and simplifying: $$(e^2 - 1)x^2 - y^2 = a^2(e^2 - 1) \iff \frac{x^2}{a^2} - \frac{y^2}{a^2(e^2 - 1)} = 1$$ Defining the conjugate semi-axis $b > 0$ by the fundamental relation: $$\mathbf{b^2 \equiv a^2(e^2 - 1) \iff e = \sqrt{1 + \frac{b^2}{a^2}} > 1}$$ yields the canonical equation of the hyperbola: $$\mathbf{\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1}$$

2. The Difference of Focal Radii

Like the ellipse, the hyperbola possesses two foci $S(ae, 0)$ and $S'(-ae, 0)$. The focal distances to any point $P(x, y)$ on the right branch are $SP = ex - a$ and $S'P = ex + a$. Their difference is: $$S'P - SP = (ex + a) - (ex - a) = \mathbf{2a = \text{Constant Everywhere!}}$$ The Focal Difference Theorem: A hyperbola is the locus of all points whose difference of distances from two fixed foci $S$ and $S'$ is constant and equal to the transverse axis: $|S'P - SP| = 2a$.

3. Canonical Geometric Elements

  • Center ($C$): $(0, 0)$.
  • Transverse Axis: Segment along the $x$-axis connecting vertices $A(a, 0)$ and $A'(-a, 0)$, length $2a$.
  • Conjugate Axis: Segment along the $y$-axis of length $2b$.
  • Foci: $S(ae, 0)$ and $S'(-ae, 0)$, distance $SS' = 2ae$.
  • Directrices: $x = \pm a/e$, distance $2a/e$.
  • Latus Rectum: Length $2b^2/a$.

Β§7.2 Asymptotes & The Conjugate Hyperbola

1. Asymptotes of the Hyperbola

An asymptote to a curve is a straight line such that the perpendicular distance from a point on the curve to the line approaches zero as the point recedes to infinity. Solving the canonical hyperbola for $y$: $$y = \pm \frac{b}{a}\sqrt{x^2 - a^2} = \pm \frac{b}{a}x \sqrt{1 - \frac{a^2}{x^2}} = \pm \frac{b}{a}x \left(1 - \frac{a^2}{2x^2} - \dots\right) \to \pm \frac{b}{a}x \quad \text{as } |x| \to \infty$$ Thus, the hyperbola possesses two real asymptotes passing through the center: $$\mathbf{y = \frac{b}{a}x \quad \text{and} \quad y = -\frac{b}{a}x \iff \frac{x^2}{a^2} - \frac{y^2}{b^2} = 0}$$ The angle between the asymptotes is $2\theta$, where $\tan \theta = b/a \implies 2\theta = 2\arctan(b/a)$.

2. The Conjugate Hyperbola

The hyperbola whose transverse and conjugate axes are respectively the conjugate and transverse axes of the given hyperbola is termed the conjugate hyperbola: $$\mathbf{\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \iff -\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1}$$ Remarkable Properties:

  • The hyperbola and its conjugate hyperbola share the exact same pair of asymptotes: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 0$.
  • If $e_1$ is the eccentricity of the original hyperbola and $e_2$ is the eccentricity of the conjugate hyperbola: $$e_1^2 = 1 + \frac{b^2}{a^2} = \frac{a^2 + b^2}{a^2} \implies \frac{1}{e_1^2} = \frac{a^2}{a^2 + b^2}$$ $$e_2^2 = 1 + \frac{a^2}{b^2} = \frac{a^2 + b^2}{b^2} \implies \frac{1}{e_2^2} = \frac{b^2}{a^2 + b^2}$$ Adding these reciprocals establishes the celebrated theorem: $$\mathbf{\frac{1}{e_1^2} + \frac{1}{e_2^2} = 1}$$

Β§7.3 Tangents, Normals & The Director Circle

1. Equations of the Tangent

  1. Point Form: Tangent at $P(x_1, y_1)$ on $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$: $$\mathbf{\frac{x x_1}{a^2} - \frac{y y_1}{b^2} = 1}$$
  2. Parametric Form: Using $x = a\sec\theta, y = b\tan\theta$: $$\mathbf{\frac{x \sec \theta}{a} - \frac{y \tan \theta}{b} = 1}$$
  3. Slope Form: A line $y = mx + c$ is tangent if and only if $c^2 = a^2 m^2 - b^2$: $$\mathbf{y = m x \pm \sqrt{a^2 m^2 - b^2} \quad (|m| > b/a)}$$

2. The Director Circle of the Hyperbola

Repeating the orthogonal tangent locus derivation for the hyperbola yields: $$\mathbf{x^2 + y^2 = a^2 - b^2}$$

  • Real Circle ($a > b$): A real concentric circle of radius $\sqrt{a^2 - b^2}$.
  • Point Circle ($a = b$): Concentrates to the origin $(0, 0)$.
  • Virtual / Imaginary ($a < b$): No pair of real mutually perpendicular tangents can be drawn to the hyperbola!

Β§7.4 The Rectangular (Equilateral) Hyperbola & xy = c^2

1. The Equilateral Hyperbola ($a = b$)

When the semi-axes are equal ($a = b$), the hyperbola is termed rectangular or equilateral: $$x^2 - y^2 = a^2$$ Key characteristics:

  • Eccentricity: $e = \sqrt{1 + a^2/a^2} = \mathbf{\sqrt{2}}$. Every rectangular hyperbola has eccentricity exactly $\sqrt{2}$!
  • Asymptotes: $y = \pm x \iff x^2 - y^2 = 0$. The asymptotes intersect at right angles ($\pi/2 = 90^\circ$).

2. Rotation of Axes to Asymptotic Coordinates ($xy = c^2$)

Because the asymptotes of a rectangular hyperbola are mutually perpendicular, they can be chosen as the coordinate axes! Rotating the axes clockwise through $45^\circ$ ($\theta = -\pi/4$): $$x = X \cos(-\pi/4) - Y \sin(-\pi/4) = \frac{X + Y}{\sqrt{2}}$$ $$y = X \sin(-\pi/4) + Y \cos(-\pi/4) = \frac{-X + Y}{\sqrt{2}}$$ Substituting into $x^2 - y^2 = a^2$: $$\left(\frac{X + Y}{\sqrt{2}}\right)^2 - \left(\frac{Y - X}{\sqrt{2}}\right)^2 = a^2 \implies \frac{(X+Y)^2 - (Y-X)^2}{2} = a^2$$ $$\frac{4XY}{2} = a^2 \implies 2XY = a^2 \iff \mathbf{XY = \frac{a^2}{2} \equiv c^2}$$ This is the widely used standard canonical form of the rectangular hyperbola: $$\mathbf{xy = c^2 \quad \left(c = \frac{a}{\sqrt{2}}\right)}$$

Β§7.5 Geometric Properties of the Asymptotic Hyperbola xy = c^2

1. Parametric Form & Tangents

Setting $x = ct$, the curve $xy = c^2$ gives $y = c/t$. The standard parametrization is: $$\mathbf{P(t) = \left(c t, \; \frac{c}{t}\right) \quad (t \ne 0)}$$ Differentiating implicitly: $y + x \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{y}{x} = -\frac{c/t}{ct} = -\frac{1}{t^2}$. The equation of the tangent at $P(t)$ is: $$y - \frac{c}{t} = -\frac{1}{t^2}(x - ct) \iff t^2 y - ct = -x + ct \iff \mathbf{x + t^2 y = 2ct \iff \frac{x}{t} + yt = 2c}$$

2. The Constant Area Tangent Triangle Theorem

The tangent at $P(t)$ intersects the asymptotes ($x$-axis and $y$-axis) at: $$A: y = 0 \implies x_A = 2ct \implies A(2ct, 0)$$ $$B: x = 0 \implies y_B = \frac{2c}{t} \implies B\left(0, \frac{2c}{t}\right)$$ Notice:

  1. Midpoint Property: The midpoint of the tangent segment $AB$ is $\left(\frac{2ct + 0}{2}, \frac{0 + 2c/t}{2}\right) = (ct, c/t) = P$. Theorem: The point of contact $P$ bisects the segment of the tangent intercepted between the asymptotes!
  2. Constant Triangle Area: The area of the right triangle $\triangle OAB$ formed by the tangent and the two coordinate asymptotes is: $$\operatorname{Area}(\triangle OAB) = \frac{1}{2} OA \cdot OB = \frac{1}{2}(2ct)\left(\frac{2c}{t}\right) = \mathbf{2c^2 = \text{Constant Everywhere!}}$$ The area is completely independent of the parameter $t$!

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 7.1: Standard Hyperbola Elements and Asymptote Equations

For the hyperbola $9x^2 - 16y^2 = 144$: (a) Find the lengths of the transverse and conjugate axes, eccentricity $e$, coordinates of the foci, and directrices. (b) Find the equations of the two asymptotes and the angle between them.

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

Transverse axis is $2a = 8$; conjugate axis is $2b = 6$.

Step 2: Eccentricity, Foci and Directrices
$$e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{9}{16}} = \frac{5}{4} = 1.25 \\ \text{Foci: } (\pm ae, 0) = \left(\pm 4\cdot\frac{5}{4}, 0\right) = (\pm 5, 0) \\ \text{Directrices: } x = \pm\frac{a}{e} = \pm\frac{4}{5/4} = \pm\frac{16}{5} = \pm 3.2$$

Eccentricity is $1.25$, foci are at $(\pm 5, 0)$, directrices are $x = \pm 3.2$.

Step 3: Asymptotes and Included Angle
$$y = \pm\frac{b}{a}x = \pm\frac{3}{4}x \iff 3x - 4y = 0 \quad \text{and} \quad 3x + 4y = 0 \\ \tan\theta = \frac{3}{4} \implies \text{Angle between asymptotes } 2\theta = 2\arctan\left(\frac{3}{4}\right) \approx 73.74^\circ$$

The asymptotes are $y = \pm \frac{3}{4}x$.

Final Answer & Physical Insight

a = 4, \; b = 3; \quad e = 1.25; \quad \text{Foci: } (\pm 5, 0); \quad \text{Directrices: } x = \pm 3.2; \quad \text{Asymptotes: } y = \pm\frac{3}{4}x

Tier 2: Intermediate Exam Example 7.2: Conjugate Hyperbolas and Eccentricity Reciprocal Theorem

A hyperbola $H_1$ has equation $16x^2 - 9y^2 = 144$. (a) Write the equation of its conjugate hyperbola $H_2$. (b) Find the eccentricities $e_1$ and $e_2$ of both hyperbolas. (c) Rigorously verify that $\frac{1}{e_1^2} + \frac{1}{e_2^2} = 1$.

Step 1: Canonical Form of $H_1$ and $H_2$
$$H_1: \frac{x^2}{9} - \frac{y^2}{16} = 1 \implies a^2 = 9, \; b^2 = 16 \\ H_2 \text{ (Conjugate)}: -\frac{x^2}{9} + \frac{y^2}{16} = 1 \iff \frac{y^2}{16} - \frac{x^2}{9} = 1 \implies 9y^2 - 16x^2 = 144$$

The conjugate hyperbola is obtained by switching signs of terms.

Step 2: Compute Eccentricities $e_1$ and $e_2$
$$e_1 = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{16}{9}} = \sqrt{\frac{25}{9}} = \frac{5}{3} \\ e_2 = \sqrt{1 + \frac{a^2}{b^2}} = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}$$

Thus $e_1 = 5/3$ and $e_2 = 5/4$.

Step 3: Verify the Reciprocal Identity
$$\frac{1}{e_1^2} + \frac{1}{e_2^2} = \frac{1}{(5/3)^2} + \frac{1}{(5/4)^2} = \frac{9}{25} + \frac{16}{25} = \frac{9 + 16}{25} = \frac{25}{25} = 1$$

The sum of reciprocals of squared eccentricities equals exactly 1.

Final Answer & Physical Insight

H_2: \frac{y^2}{16} - \frac{x^2}{9} = 1; \quad e_1 = \frac{5}{3}, \; e_2 = \frac{5}{4}; \quad \frac{1}{e_1^2} + \frac{1}{e_2^2} = \frac{9}{25} + \frac{16}{25} = 1

Tier 3: Honors / Proof Challenge Example 7.3: Rectangular Hyperbola $xy = c^2$: Concurrency of Circle Intersections

A circle $x^2 + y^2 + 2gx + 2fy + k = 0$ intersects the rectangular hyperbola $xy = c^2$ in four points $P_i(x_i, y_i)$ with parameters $t_i$ ($i = 1, 2, 3, 4$). (a) Derive the quartic equation in $t$. (b) Prove that $t_1 t_2 t_3 t_4 = 1$. (c) Prove that the product of the four abscissae is $c^4$, the product of the four ordinates is $c^4$, and the center of mean position of the four points is $(-g/2, -f/2)$.

Step 1: Substitute Parametric Coordinates into Circle
$$x = ct, \quad y = \frac{c}{t} \\ (ct)^2 + \left(\frac{c}{t}\right)^2 + 2g(ct) + 2f\left(\frac{c}{t}\right) + k = 0 \\ c^2 t^2 + \frac{c^2}{t^2} + 2gct + \frac{2fc}{t} + k = 0$$

Multiply by $t^2$ to clear the denominator.

Step 2: Form the Quartic Polynomial in $t$
$$c^2 t^4 + 2gc t^3 + k t^2 + 2fc t + c^2 = 0$$

Dividing by $c^2$: $t^4 + \frac{2g}{c}t^3 + \frac{k}{c^2}t^2 + \frac{2f}{c}t + 1 = 0$.

Step 3: Apply Viète's Formulas
$$\sum t_i = t_1 + t_2 + t_3 + t_4 = -\frac{2g}{c} \\ \sum t_i t_j t_k = -\frac{2f}{c} \\ t_1 t_2 t_3 t_4 = \frac{c^2}{c^2} = 1$$

The product of the parameters is identically $t_1 t_2 t_3 t_4 = 1$.

Step 4: Prove Coordinate Products and Centroid
$$x_1 x_2 x_3 x_4 = (ct_1)(ct_2)(ct_3)(ct_4) = c^4(t_1 t_2 t_3 t_4) = c^4(1) = c^4 \\ y_1 y_2 y_3 y_4 = \left(\frac{c}{t_1}\right)\left(\frac{c}{t_2}\right)\left(\frac{c}{t_3}\right)\left(\frac{c}{t_4}\right) = \frac{c^4}{t_1 t_2 t_3 t_4} = c^4 \\ \bar{x} = \frac{\sum x_i}{4} = \frac{c \sum t_i}{4} = \frac{c(-2g/c)}{4} = -\frac{g}{2}, \quad \bar{y} = \frac{\sum y_i}{4} = \frac{c \sum(1/t_i)}{4} = -\frac{f}{2}$$

The centroid of the four intersection points is $(-g/2, -f/2)$, which is the exact midpoint of the line joining the origin to the circle's center $(-g, -f)$.

Final Answer & Physical Insight

t_1 t_2 t_3 t_4 = 1; \quad \prod x_i = c^4, \; \prod y_i = c^4; \quad \text{Centroid: } \left(-\frac{g}{2}, -\frac{f}{2}\right)