Hyperbola in Coordinate Geometry

Hyperbola: Definition (Difference of Distances from Two Foci)

The hyperbola is another type of non-degenerate conic section, formed when a plane intersects both nappes of a double cone. Like the ellipse, the hyperbola can also be defined as a locus of points using distances to two fixed points called foci, but with a key difference: the constant is the absolute difference of the distances, not the sum.

Definition (Based on Foci)

A hyperbola is defined as the locus of all points $P$ in a plane such that the absolute difference of the distances from $P$ to two distinct fixed points, $F_1$ and $F_2$, in the plane is a constant positive value. The two distinct fixed points $F_1$ and $F_2$ are called the foci of the hyperbola.

Let $P(x, y)$ be any point on the hyperbola.

Let $F_1$ and $F_2$ be the two fixed foci.

The defining condition for any point $P$ to be on the hyperbola is:

$\mathbf{||PF_1| - |PF_2|| = \text{Constant}}$

Let the constant absolute difference be $2a$. The condition becomes $||PF_1| - |PF_2|| = 2a$.

The constant difference ($2a$) must be positive and strictly less than the distance between the two foci ($|F_1 F_2|$). So, $0 < 2a < |F_1 F_2|$, which implies $0 < a < c$, where $2c = |F_1 F_2|$ is the distance between the foci.

The hyperbola consists of two separate, unbounded curves called branches. For any point P on one branch, $|PF_1| - |PF_2| = 2a$, and for any point P on the other branch, $|PF_2| - |PF_1| = 2a$. The absolute value combines both cases.

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Key Terminology Associated with Foci Definition

Based on the foci definition, we can identify and define several key geometric features of a hyperbola:

• Foci ($F_1, F_2$): The two fixed points used in the definition ($||PF_1| - |PF_2|| = \text{constant}$).

• Center (C): The midpoint of the line segment joining the two foci ($F_1 F_2$). The hyperbola is symmetric about its center.

• Transverse Axis: The line segment that passes through the two foci and has its endpoints on the two branches of the hyperbola. The length of the transverse axis is equal to the constant absolute difference of distances, $2a$.

• Vertices ($V_1, V_2$): The two endpoints of the transverse axis. They are the points where the hyperbola intersects the line passing through the foci. The distance from the center to each vertex is $a$. The vertices are the points on the hyperbola closest to each other.

• Conjugate Axis: The line segment that passes through the center, is perpendicular to the transverse axis. Its length is denoted by $2b$, and $b$ is related to $a$ and $c$ (distance from center to focus) by the relationship $c^2 = a^2 + b^2$. The endpoints of the conjugate axis are sometimes called the co-vertices.

The relationship $c^2 = a^2 + b^2$ is a key identity for hyperbolas, linking the distance from the center to the focus ($c$), the semi-transverse axis ($a$), and the semi-conjugate axis ($b$). Note that for a hyperbola, $c > a$, which is consistent with $c^2 = a^2 + b^2$ when $b>0$. Also, $a$ is NOT necessarily greater than $b$ in a hyperbola.

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Standard Equations of Hyperbola ($\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, etc.)

The algebraic equations for a hyperbola are derived using its definition based on the absolute difference of distances to two foci. The standard forms of these equations are obtained when the center of the hyperbola is at the origin $(0, 0)$ and its transverse axis (the axis containing the foci and vertices) lies along either the x-axis or the y-axis.

Case 1: Foci on the x-axis (Transverse Axis along x-axis)

Let the center of the hyperbola be at the origin $C(0, 0)$.

Let the foci be located on the x-axis, symmetric with respect to the origin. Let their coordinates be $F_1(-c, 0)$ and $F_2(c, 0)$, where $c$ is a positive constant representing the distance from the center to each focus ($c > 0$). The distance between the foci is $2c$.

By the definition of a hyperbola, for any point $P(x, y)$ on the hyperbola, the absolute difference of the distances from $P$ to the foci is a constant value, which we denote as $2a$. So, $||PF_1| - |PF_2|| = 2a$. For a hyperbola to be formed, the constant difference $2a$ must be strictly less than the distance between the foci $2c$. Thus, $0 < 2a < 2c$, which implies $0 < a < c$.

Using the distance formula for $|PF_1|$ and $|PF_2|$:

$|PF_1| = \sqrt{(x - (-c))^2 + (y - 0)^2} = \sqrt{(x + c)^2 + y^2}$

$|PF_2| = \sqrt{(x - c)^2 + (y - 0)^2} = \sqrt{(x - c)^2 + y^2}$

Substitute these into the hyperbola definition equation $||PF_1| - |PF_2|| = 2a$:

Removing the absolute value, we get $\sqrt{(x + c)^2 + y^2} - \sqrt{(x - c)^2 + y^2} = \pm 2a$.

Derivation Steps (Algebraic Manipulation):

Similar to the ellipse derivation, but with a difference and absolute value, the process involves isolating radicals and squaring multiple times. Starting from $\sqrt{(x + c)^2 + y^2} - \sqrt{(x - c)^2 + y^2} = \pm 2a$:

1. Isolate one radical:

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

2. Square both sides:

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

   $x^2 + 2cx + c^2 + y^2 = 4a^2 \pm 4a\sqrt{(x - c)^2 + y^2} + x^2 - 2cx + c^2 + y^2$

3. Simplify by canceling terms ($x^2, c^2, y^2$):

   $2cx = 4a^2 \pm 4a\sqrt{(x - c)^2 + y^2} - 2cx$

   $4cx - 4a^2 = \pm 4a\sqrt{(x - c)^2 + y^2}$

4. Divide by 4:

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

5. Square both sides again:

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

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

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

6. Simplify by canceling $-2a^2cx$:

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

7. Rearrange terms: group x terms, y terms, and constants:

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

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

8. Define a new constant $b^2 = c^2 - a^2$. Since $c > a > 0$, we have $c^2 > a^2$, so $c^2 - a^2 > 0$ and $b^2$ is positive.

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

   (Substitute $c^2 - a^2 = b^2$)

9. Divide the entire equation by $a^2 b^2$ (assuming $a \neq 0, b \neq 0$):

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

   Simplify:

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

   (Standard form 1)

This is the standard equation of a hyperbola centered at the origin with foci on the x-axis. The transverse axis lies along the x-axis. Its length is $2a$, and the vertices are $(\pm a, 0)$. The conjugate axis lies along the y-axis. Its length is $2b$. The foci are at $(\pm c, 0) = (\pm \sqrt{a^2 + b^2}, 0)$. The relationship $c^2 = a^2 + b^2$ holds, where $a$ is related to the vertices (transverse axis) and $b$ is related to the conjugate axis.

The vertices $(\pm a, 0)$ satisfy the equation: $\frac{(\pm a)^2}{a^2} - \frac{0^2}{b^2} = \frac{a^2}{a^2} - 0 = 1$.

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Case 2: Foci on the y-axis (Transverse Axis along y-axis)

If the foci are located on the y-axis, symmetric with respect to the origin, at $F_1(0, -c)$ and $F_2(0, c)$, and the constant absolute difference of distances is $2a$ ($0 < a < c$), a similar derivation leads to the standard equation for a hyperbola opening upwards and downwards.

The equation $||PF_1| - |PF_2|| = 2a$ becomes $|\sqrt{(x - 0)^2 + (y - (-c))^2} - \sqrt{(x - 0)^2 + (y - c)^2}| = 2a$, or $|\sqrt{x^2 + (y + c)^2} - \sqrt{x^2 + (y - c)^2}| = 2a$.

Performing analogous algebraic manipulations, substituting $b^2 = c^2 - a^2$, and rearranging, we arrive at the standard equation for a hyperbola centered at the origin with foci on the y-axis:

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

In this equation, the variable with the positive squared term ($y^2$) indicates the direction of the transverse axis (y-axis). The denominator under the positive term is $a^2$, where $a$ is the distance from the center to each vertex. So, the vertices are $(0, \pm a)$, and the length of the transverse axis is $2a$. The denominator under the negative term is $b^2$, where $b$ is related to the conjugate axis, length $2b$. The relationship $c^2 = a^2 + b^2$ still holds, where $c$ is the distance from the center to the focus. The foci are at $(0, \pm c) = (0, \pm \sqrt{a^2 + b^2})$.

Key Distinction Between the Two Standard Forms:

For standard hyperbolas centered at the origin, the form is $\frac{\text{variable}^2}{\text{denominator}_1} - \frac{\text{variable}^2}{\text{denominator}_2} = 1$. The variable with the positive squared term determines the location of the transverse axis and vertices.

• If $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the $x^2$ term is positive. The transverse axis is horizontal (along the x-axis). The vertices are at $(\pm a, 0)$. $a^2$ is under $x^2$.
• If $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$, the $y^2$ term is positive. The transverse axis is vertical (along the y-axis). The vertices are at $(0, \pm a)$. $a^2$ is under $y^2$.

In hyperbolas, $a$ is always the semi-transverse axis length (distance from center to vertex), which is the denominator of the positive term. $b$ is always the semi-conjugate axis length. There is no requirement for $a$ to be greater than $b$ (unlike ellipses).

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Answer:

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Properties of Hyperbola (Foci, Vertices, Axes, Eccentricity, Directrices, Latus Rectum, Asymptotes)

Just as with parabolas and ellipses, the standard equations of hyperbolas centered at the origin allow us to systematically identify their key geometric properties. These properties define the shape, orientation, and significant points and lines associated with the hyperbola.

The standard equations for a hyperbola centered at the origin are $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ or $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$. In these equations, $a$ is always the distance from the center to a vertex (semi-transverse axis length), and $b$ is the semi-conjugate axis length ($a>0, b>0$). The distance from the center to each focus is $c$, related by $c^2 = a^2 + b^2$, with $c > a > 0$.

Eccentricity ($e$)

The eccentricity of a hyperbola is defined as the ratio of the distance from the center to a focus ($c$) to the distance from the center to a vertex ($a$).

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

Since $c^2 = a^2 + b^2$ and $a, b > 0$, we have $c > a$. Therefore, the eccentricity $e$ of a hyperbola always satisfies $e > 1$.

We can express $c$ and $b^2$ in terms of $a$ and $e$ using $c = ae$:

• $\mathbf{c = ae}$
• $b^2 = c^2 - a^2 = (ae)^2 - a^2 = a^2 e^2 - a^2 = a^2(e^2 - 1)$. Since $e > 1$, $e^2 - 1 > 0$, so $b^2$ is positive.

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Directrices

Similar to ellipses, hyperbolas also have two directrices, one for each focus, defined by the property $|PF| = e |PM|$, where $e > 1$. The directrices are lines perpendicular to the transverse axis and are symmetric with respect to the center.

• For the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ (transverse axis on x-axis, foci at $(\pm c, 0)$):

  The equations of the two directrices are $\mathbf{x = \pm \frac{a}{e}}$. The directrix $x = a/e$ corresponds to the focus $(c, 0) = (ae, 0)$, and the directrix $x = -a/e$ corresponds to the focus $(-c, 0) = (-ae, 0)$. Since $e > 1$, $a/e < a$, confirming that the directrices are vertical lines located between the vertices and the center.

• For the hyperbola $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$ (transverse axis on y-axis, foci at $(0, \pm c)$):

  The equations of the two directrices are $\mathbf{y = \pm \frac{a}{e}}$. The directrix $y = a/e$ corresponds to the focus $(0, c) = (0, ae)$, and the directrix $y = -a/e$ corresponds to the focus $(0, -c) = (0, -ae)$. These are horizontal lines located between the vertices and the center.

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Latus Rectum

The latus rectum of a hyperbola is a chord passing through one of the foci, perpendicular to the transverse axis, and having its endpoints on the hyperbola. A hyperbola has two latera recta, one for each focus.

Length of the Latus Rectum:

Consider the standard hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ (transverse axis on x-axis), centered at the origin with foci at $(\pm c, 0)$. The latus rectum through the focus $F_2(c, 0)$ is a vertical line with the equation $x = c$.

To find the y-coordinates of the endpoints, substitute $x = c$ into the equation of the hyperbola:

Solve equation (ii) for $y^2$:

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

From the relationship $b^2 = c^2 - a^2$, we can substitute $c^2 - a^2$ with $b^2$:

Solve equation (iii) for $y^2$:

Take the square root of equation (iv) to find the y-coordinates:

The endpoints of the latus rectum through $F_2(c, 0)$ are $(c, b^2/a)$ and $(c, -b^2/a)$.

The length of the latus rectum is the distance between these two endpoints. Since they have the same x-coordinate, the distance is the absolute difference of their y-coordinates:

Since $a > 0$ and $b > 0$, $\frac{2b^2}{a}$ is positive.

This length is the same for the other latus rectum (through $F_1$) and also for the standard hyperbola with vertical transverse axis $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$.

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Asymptotes

A unique characteristic of hyperbolas is the presence of two straight lines called asymptotes. The branches of the hyperbola get arbitrarily close to these lines as they extend towards infinity, but they never actually touch them. The asymptotes pass through the center of the hyperbola.

For the standard hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ (transverse axis on x-axis, center at origin), the equations of the asymptotes can be found by setting the right-hand side of the equation to zero and solving for $y$:

$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 0$

Rearrange the terms:

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

Multiply by $b^2$:

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

Take the square root of both sides:

For the standard hyperbola $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$ (transverse axis on y-axis, center at origin), setting the RHS to zero and solving for $y$:

$\frac{y^2}{a^2} - \frac{x^2}{b^2} = 0$

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

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

Take the square root:

The slopes of the asymptotes are $\pm \frac{b}{a}$ or $\pm \frac{a}{b}$, depending on the orientation. They can be visualized as the diagonals of the rectangle formed by lines $x=\pm a, y=\pm b$ (for horizontal transverse axis) or $x=\pm b, y=\pm a$ (for vertical transverse axis). This rectangle is called the conjugate rectangle or the fundamental rectangle.

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Summary of Properties (Hyperbola Centered at Origin)

For standard hyperbolas centered at the origin $(0, 0)$, with semi-transverse axis $a$ and semi-conjugate axis $b$ (where $a>0, b>0$). The distance from the center to each focus is $c = \sqrt{a^2 + b^2}$. Eccentricity $e = c/a$.

Property	$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$	$\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$
(Common: Center $(0,0)$, $c=\sqrt{a^2+b^2}$, $e=c/a > 1$)	(Transverse Axis on x-axis)	(Transverse Axis on y-axis)
Center	$(0, 0)$	$(0, 0)$
Transverse Axis Location	Along x-axis ($y=0$)	Along y-axis ($x=0$)
Length of Transverse Axis	$2a$	$2a$
Length of Semi-transverse Axis	$a$	$a$
Length of Conjugate Axis	$2b$	$2b$
Length of Semi-conjugate Axis	$b$	$b$
Vertices	$(\pm a, 0)$	$(0, \pm a)$
Co-vertices (Endpoints of Conjugate Axis)	$(0, \pm b)$	$(\pm b, 0)$
Foci	$(\pm c, 0) = (\pm ae, 0)$	$(0, \pm c) = (0, \pm ae)$
Eccentricity ($e = c/a$)	$e = \frac{\sqrt{a^2+b^2}}{a}$	$e = \frac{\sqrt{a^2+b^2}}{a}$
Equations of Directrices	$x = \pm \frac{a}{e}$	$y = \pm \frac{a}{e}$
Length of Latus Rectum	$\frac{2b^2}{a}$	$\frac{2b^2}{a}$
Equations of Asymptotes	$y = \pm \frac{b}{a} x$	$y = \pm \frac{a}{b} x$

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Answer:

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Parametric Equations of Hyperbola

Similar to other conic sections, hyperbolas can also be represented using parametric equations, expressing the coordinates $(x, y)$ of points on the curve as functions of a single parameter. This representation is derived using trigonometric or hyperbolic function identities that resemble the structure of the hyperbola's Cartesian equation.

Parametric Form using Hyperbolic Functions

The standard hyperbolic identity is $\cosh^2 t - \sinh^2 t = 1$, where $\cosh t = \frac{e^t + e^{-t}}{2}$ and $\sinh t = \frac{e^t - e^{-t}}{2}$. This identity directly mirrors the structure of the standard hyperbola equations (difference of two squares equals 1).

Consider the standard hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ (transverse axis on x-axis, center at origin). We can set $\frac{x}{a} = \cosh t$ and $\frac{y}{b} = \sinh t$.

Substitute $x = a \cosh t$ and $y = b \sinh t$ into the hyperbola equation:

Simplify equation (i):

Equation (iii) is the standard hyperbolic identity, which is true for all real values of $t$. Thus, the point $(a \cosh t, b \sinh t)$ lies on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ for any real $t$.

The parametric equations using hyperbolic functions for $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ are:

$\mathbf{x = a \cosh t}$

$\mathbf{y = b \sinh t}$

where $t$ is any real parameter. As $t$ varies from $-\infty$ to $\infty$, $\cosh t$ takes values in $[1, \infty)$. Thus, $x = a \cosh t$ takes values in $[a, \infty)$ (for $a>0$), tracing only the right branch of the hyperbola. To include the left branch, we can use $x = -a \cosh t$ and $y = b \sinh t$, which gives the branch where $x \le -a$. Alternatively, some definitions use $x = \pm a \cosh t, y = b \sinh t$ for both branches.

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Parametric Form using Trigonometric Functions

We can also use trigonometric functions, specifically secant and tangent, due to the identity $\sec^2 \theta - \tan^2 \theta = 1$. This identity also fits the hyperbola equation structure.

For the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ (transverse axis on x-axis, center at origin):

Set $\frac{x}{a} = \sec \theta$ and $\frac{y}{b} = \tan \theta$.

Substitute $x = a \sec \theta$ and $y = b \tan \theta$ into the hyperbola equation:

Simplify equation (iv):

Equation (vi) is the standard trigonometric identity, which is true for values of $\theta$ where $\sec \theta$ and $\tan \theta$ are defined (i.e., $\theta \neq \frac{\pi}{2} + n\pi$ for integer $n$). The parametric equations using trigonometric functions are:

$\mathbf{x = a \sec \theta}$

$\mathbf{y = b \tan \theta}$

where $\theta$ is the parameter. As $\theta$ varies (excluding $\pi/2, 3\pi/2$, etc.), this traces both branches of the hyperbola. For instance, as $\theta \to \pi/2^-$, $\sec \theta \to +\infty$, $\tan \theta \to +\infty$, $x \to +\infty$, $y \to +\infty$ (upper right branch). As $\theta \to \pi/2^+$, $\sec \theta \to -\infty$, $\tan \theta \to -\infty$, $x \to -\infty$, $y \to -\infty$ (lower left branch).

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

For the hyperbola with the transverse axis along the y-axis, $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$ (center at origin), the parametric equations are similarly derived using the same identities. The key is to match the positive term with the squared term in the identity.

• Using hyperbolic functions ($\cosh^2 t - \sinh^2 t = 1$):

  Set $\frac{y}{a} = \cosh t$ and $\frac{x}{b} = \sinh t$. Substitute $x = b \sinh t$ and $y = a \cosh t$ into the equation: $\frac{(a \cosh t)^2}{a^2} - \frac{(b \sinh t)^2}{b^2} = \cosh^2 t - \sinh^2 t = 1$.

  $\mathbf{x = b \sinh t, \quad y = a \cosh t}$

  As $t$ varies, $\cosh t \ge 1$, so $y = a \cosh t$ traces values in $[a, \infty)$ (for $a>0$), tracing only the upper branch. For the lower branch, use $y = -a \cosh t$ and $x = b \sinh t$, giving $y \le -a$.

• Using trigonometric functions ($\sec^2 \theta - \tan^2 \theta = 1$):

  Set $\frac{y}{a} = \sec \theta$ and $\frac{x}{b} = \tan \theta$. Substitute $x = b \tan \theta$ and $y = a \sec \theta$ into the equation: $\frac{(a \sec \theta)^2}{a^2} - \frac{(b \tan \theta)^2}{b^2} = \sec^2 \theta - \tan^2 \theta = 1$.

  $\mathbf{x = b \tan \theta, \quad y = a \sec \theta}$

  As $\theta$ varies (excluding $\pi/2, 3\pi/2$, etc.), this traces both branches of the hyperbola.

Parametric Form for General Hyperbola (Center $(h, k)$):

If the center is at $(h, k)$, translate the standard parametric equations. For example, for $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$ (transverse axis horizontal), the parametric equations are:

$\frac{x-h}{a} = \sec \theta \implies \mathbf{x = h + a \sec \theta}$

$\frac{y-k}{b} = \tan \theta \implies \mathbf{y = k + b \tan \theta}$

Using hyperbolic functions: $\mathbf{x = h + a \cosh t, \quad y = k + b \sinh t}$ (for the right branch).

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Answer:

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