Problem 55
Question
In Exercises 51-56, sketch (if possible) the graph of the degenerate conic. \(x^2+2xy+y^2-1=0\)
Step-by-Step Solution
Verified Answer
The graph of the conic is a circle centered at the origin with a radius of 1.
1Step 1: Rewrite the Equation
The given equation is not in any standard form for a conic. Hence, to visualize the graph of this particular equation, it would be very beneficial to convert it to a recognizable form. After completing the square, the equation remains the same as \(x^2+2xy+y^2-1=0\). Now this can be rewritten as \( (x+y)^2 = 1 \).
2Step 2: Identify the Conic
The equation \((x+y)^2=1\) is the formula for a circle centered around the origin with a radius of 1. The sign of both the x and the y variables are the same and thus it is a circle.
3Step 3: Sketch the Graph
Draw two axes, label them x and y. Now plot a circle centered at the origin (0,0) with radius 1. The degenerate conic is a circle.
Key Concepts
Conic SectionsGraphing EquationsCompleting the Square
Conic Sections
Conic sections are the curves obtained by intersecting a cone with a plane in different angles. They are fundamental objects in geometry and include:
However, equations aren't always presented in these standard forms. Sometimes they appear as more complex expressions as seen in the exercise. When certain conditions are met, these conics can also become "degenerate," which occurs when the conic section collapses into a simpler form—such as a point or a line. In the case of this exercise, we are dealing with a degenerate case, showing how even non-standard equations can represent common conic shapes like circles.
- Circles
- Ellipses
- Parabolas
- Hyperbolas
However, equations aren't always presented in these standard forms. Sometimes they appear as more complex expressions as seen in the exercise. When certain conditions are met, these conics can also become "degenerate," which occurs when the conic section collapses into a simpler form—such as a point or a line. In the case of this exercise, we are dealing with a degenerate case, showing how even non-standard equations can represent common conic shapes like circles.
Graphing Equations
Graphing equations, especially those involving conic sections, requires understanding their equations and recognizing the form they represent. The goal is to convert complex equations into a simpler, recognizable form that can easily be sketched on a graph.
For the equation given in the exercise, \(x^2 + 2xy + y^2 - 1 = 0\), the complexity initially masks the geometric shape it represents.
For the equation given in the exercise, \(x^2 + 2xy + y^2 - 1 = 0\), the complexity initially masks the geometric shape it represents.
- The process involves analyzing the terms with \(x \) and \(y\) and manipulating them using algebraic methods.
- By simplifying the equation via completing the square, it becomes possible to identify the conic section.
- Ultimately, the equation becomes \( (x+y)^2 = 1\), revealing the underlying shape as a circle.
Completing the Square
Completing the square is a valuable algebraic technique used to simplify quadratic expressions. It's particularly useful in graphing equations to bring out recognizable forms, especially with conic sections.
- The goal is to manipulate an equation into a perfect square trinomial, which is easily identifiable with specific conics.
- In the exercise, the original equation \(x^2 + 2xy + y^2 - 1 = 0\) was rewritten as \( (x+y)^2 = 1\).
- This is done by recognizing patterns within the equation and rearranging terms so that they fit the pattern of a square binomial.
Other exercises in this chapter
Problem 55
In Exercises 55-64, use a graphing utility to find one set of polar coordinates for the point given in rectangular coordinates. \(\left(3, -2\right)\)
View solution Problem 55
In Exercises 49-56, use a graphing utility to graph the curve represented by the parametric equations. Witch of Agnesi: \(\quad x= 2 \cot\ \theta, \quad y= 2 \s
View solution Problem 55
In Exercises 51-60, find the standard form of the equation of the parabola with the given characteristics. Vertex: \((4, 3) \quad\) focus: \((6, 3)\)
View solution Problem 55
In Exercises 51-58, find the distance between the point and the line. \(\textit{Point}\) \((6, 2)\) \(\textit{Line}\) \(x + 1 = 0\)
View solution