Problem 22
Question
The given curve is part of the graph of an equation in \(x\) and \(y .\) Find the equation by eliminating the parameter. $$x=t^{2}+1, \quad y=t^{2}-1, \quad \text { any real number } t$$
Step-by-Step Solution
Verified Answer
Question: Eliminate the parameter t from the given parametric equations: x = t^2 + 1, y = t^2 - 1.
Answer: The equation without the parameter t is y = x - 2.
1Step 1: Choose one equation to solve for t
We can start by solving x = t^2 + 1 for t. We pick this equation because it’s easier to isolate t compared to the y equation.
2Step 2: Solve for t
We can isolate t in the x equation as follows:
\(t^2 = x - 1.\)
Taking the square root of both sides, we have:
$$t = \pm\sqrt{x-1}.$$
3Step 3: Substitute the expression of t into the other equation
Now, substitute the expression of t we found in the previous step into the y equation:
$$y = (\pm\sqrt{x-1})^2 -1.$$
4Step 4: Simplify the equation
Simplify the equation to obtain the curve equation without the parameter t:
$$y = (x - 1) - 1,$$
which gives us the final equation in terms of x and y:
$$y = x - 2.$$
Key Concepts
Eliminating ParametersCoordinate GeometryAlgebraic Manipulation
Eliminating Parameters
Eliminating parameters involves removing the parameter "t" from a set of parametric equations to find a direct relationship between variables, often denoted as "x" and "y".
In parametric equations, each variable is expressed in terms of a parameter. The goal is to simplify to the conventional format of an equation in coordinate geometry, usually without the parameter.
When given parametric equations:
- Equation for "x": relates x with the parameter t.
- Equation for "y": relates y with the same parameter t.
Coordinate Geometry
Coordinate geometry, also known as analytic geometry, describes geometric figures using a coordinate system, typically the Cartesian coordinate system. Here, geometric shapes are expressed in terms of algebraic equations.
Converting parametric equations into standard x-y equations allows for easier understanding and plotting of curves on the Cartesian plane. Adjusting and simplifying these connections through algebra helps illustrate how shapes and lines relate to each other in space.
After removing the parameter with algebraic manipulation, you are often left with a simple, recognizable form.
- Recognizable equations, such as y = mx + c, represent a straight line.
- More complex shapes, like parabolas or circles, can also be expressed this way.
Algebraic Manipulation
Algebraic manipulation involves working with equations to simplify the expressions, rearrange terms, or alter the form to achieve the desired result.
When eliminating parameters, the core action relies on algebra to isolate variables and substitute the parameter.
In our exercise, algebraic manipulation was crucial in steps:
- First, solve one equation for the parameter t, producing t = ±√(x-1).
- Then, substitute this into the equation for y, giving y = (±√(x-1))² - 1.
- The simplification of y = (x - 1) - 1 finally yields y = x - 2.
Other exercises in this chapter
Problem 21
In Exercises \(17-28,\) determine the vertex, focus, and directrix of the parabola without graphing and state whether it opens upward, downward, left, or right.
View solution Problem 21
Find the equation of the hyperbola that satisfies the given conditions. Center (0,0)\(; x\) -intercepts \(\pm 3 ;\) asymptote \(y=2 x\)
View solution Problem 22
Convert the rectangular coordinates to polar coordinates. $$(2 \sqrt{3}, 2)$$
View solution Problem 22
Sketch the graph of the equation and label the vertices. $$r=\frac{5}{3+2 \sin \theta}$$
View solution