Question

Eliminate the parameter in the given parametric equations. Describe the curve defined by the parametric equations based on its rectangular form. \(x=a \sec t+h, \quad y=b \tan t+k\)

Short answer

The curve is a hyperbola, centered at (h, k).

Step-by-step solution

Express the Trig Function

We start with the given parametric equations: \(x = a \sec t + h\) and \(y = b \tan t + k\). First, isolate the trigonometric functions by subtracting the constants: \(x - h = a \sec t\) and \(y - k = b \tan t\).

Relate Secant and Tangent

Recall the trigonometric identity \(\sec^2 t = 1 + \tan^2 t\). Solve for \(\sec t\) using \(x - h = a \sec t\): \(\sec t = \frac{x-h}{a}\).

Substitute and Simplify

Substitute \(\sec t = \frac{x-h}{a}\) into the identity \(\sec^2 t = 1 + \tan^2 t\): \[\left(\frac{x-h}{a}\right)^2 = 1 + \left(\frac{y-k}{b}\right)^2\].

Convert to Rectangular Form

Rearrange the above equation to eliminate the parameter \(t\): \[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\].

Identify the Curve Type

The equation \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \) represents a hyperbola centered at (h, k). The transverse axis is horizontal.

Key concepts

Rectangular Form of Curves

In parametric equations, curves are traditionally expressed as separate functions for `x` and `y` in terms of a parameter `t`. These are known as parametric forms. However, to understand and analyze the curve as a whole, it is useful to convert these into a single equation in terms of `x` and `y`; this is known as the rectangular form.
For example, given the parametric equations:

• \(x = a \sec t + h\)
• \(y = b \tan t + k\)

Our task is to eliminate the parameter `t` to write this as a singular equation involving only `x` and `y`.
This conversion allows us to readily identify the type of curve, its properties, and the region it occupies in the coordinate plane.

Hyperbola

Hyperbolas are a type of conic section, much like circles, ellipses, and parabolas. They are defined by a specific rectangular equation: \[\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\] where (h, k) is the center of the hyperbola.
In the equation derived from parametric form, the appearance of a minus sign clarifies that we are dealing with a hyperbola rather than an ellipse.

• The term \(\frac{(x-h)^2}{a^2}\) corresponds to the transverse axis, which is aligned horizontally when the `x` term comes first.
• Similarly, \(\frac{(y-k)^2}{b^2}\) relates to the conjugate axis in the hyperbola definition.

Understanding these components not only helps identify the graph shape but also provides insights into its dimensions and orientation.

Trigonometric Identities

Identities like \(\sec^2 t = 1 + \tan^2 t\) are pivotal in manipulating and solving trigonometric equations. These identities help to link different trigonometric functions and can simplify complex expressions.
In the given problem:

• Knowing \(\sec t\) and \(\tan t\) are connected through \(\sec^2 t = 1 + \tan^2 t\) allows us to substitute and derive a simplified rectangular equation.

This identity is essential in transforming the expression to a form where the parameter `t` can be removed, bridging the gap between the parametric and rectangular forms.
Without such identities, solving these types of problems would be much more complicated, requiring significantly more steps.

Eliminating the Parameter

Eliminating the parameter in a set of parametric equations is a critical step in converting them to a rectangular form. By expressing the parameter `t` in terms of one variable, then substituting in the second equation, we eliminate the parameter.

• For \( x = a \sec t + h \), rearrange to get \( \sec t = \frac{x-h}{a} \).
• Use the identity \( \sec^2 t = 1 + \tan^2 t \) to substitute \( \sec t \) in terms of \( \tan t \).
• This substitution results in an equation connecting `x` and `y` directly.

The process of eliminating the parameter simplifies the problem, allowing for easier interpretation and application of various types of geometric analysis on the curve.