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
- Point Form: Tangent at $P(x_1, y_1)$ on $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$: $$\mathbf{\frac{x x_1}{a^2} - \frac{y y_1}{b^2} = 1}$$
- Parametric Form: Using $x = a\sec\theta, y = b\tan\theta$: $$\mathbf{\frac{x \sec \theta}{a} - \frac{y \tan \theta}{b} = 1}$$
- 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:
- 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!
- 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 $t$!
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.
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.
Transverse axis is $2a = 8$; conjugate axis is $2b = 6$.
Eccentricity is $1.25$, foci are at $(\pm 5, 0)$, directrices are $x = \pm 3.2$.
The asymptotes are $y = \pm \frac{3}{4}x$.
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
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 $\frac{1}{e_1^2} + \frac{1}{e_2^2} = 1$.
The conjugate hyperbola is obtained by switching signs of terms.
Thus $e_1 = 5/3$ and $e_2 = 5/4$.
The sum of reciprocals of squared eccentricities equals exactly 1.
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
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 $(-g/2, -f/2)$.
Multiply by $t^2$ to clear the denominator.
Dividing by $c^2$: $t^4 + \frac{2g}{c}t^3 + \frac{k}{c^2}t^2 + \frac{2f}{c}t + 1 = 0$.
The product of the parameters is identically $t_1 t_2 t_3 t_4 = 1$.
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 $(-g, -f)$.
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)