Problem 61
Question
In Exercises 57-72, classify the graph of the equation as a circle, a parabola, an ellipse, or a hyperbola. \(y^2-4x^2+4x-2y-4=0\)
Step-by-Step Solution
Verified Answer
The graph of the equation is a Circle.
1Step 1: Rearrange the equation
First step is to rearrange the equation so that we have the terms involving y together and the terms involving x together: \(y^2+2y=4x^2-4x+4\). This will prepare the equation to be complete the square.
2Step 2: Complete the square.
Complete the square on both y and x terms. For y, we have \(y^2+2y+1\) - the constant term needed for a perfect square is 1. For x, we have \(4x^2-4x+1\), the constant term needed is 1/4. The equation becomes: \((y+1)^2 = (2x-1)^2+3\).
3Step 3: Classify the graph of the equation
Now, the equation is in the standard form of a circle, ellipse, parabola or a hyperbola. The given equation is in the form of \((y-k)^2 = (x-h)^2 + r^2\), this is the form of a Circle. Therefore, the graph of the equation is a Circle.
Key Concepts
Completing the SquareConic SectionsStandard Form of a Circle
Completing the Square
When we want to solve quadratic equations or rearrange equations of conic sections, one technique we can use is completing the square. Completing the square involves rewriting a quadratic expression as a perfect square trinomial plus or minus a constant. To do this, we take the coefficient of the linear term, divide it by 2, square it, and then add and subtract this number within the expression.
This method proves particularly useful in conic sections, as it allows us to transform the equation into standard form, making it easier to identify whether we're dealing with a circle, an ellipse, a parabola, or a hyperbola. In the exercise provided, we used completing the square for both the x-terms and y-terms separately. It's important to always balance the equation by adding the same value to both sides after completing the square. This maintains the equality and prepares the equation for further classification.
This method proves particularly useful in conic sections, as it allows us to transform the equation into standard form, making it easier to identify whether we're dealing with a circle, an ellipse, a parabola, or a hyperbola. In the exercise provided, we used completing the square for both the x-terms and y-terms separately. It's important to always balance the equation by adding the same value to both sides after completing the square. This maintains the equality and prepares the equation for further classification.
Conic Sections
Conic sections are curves obtained by intersecting a cone with a plane at different angles. There are four basic types of conic sections:
- Circles: Formed when the cutting plane is perpendicular to the cone's axis.
- Ellipses: Occur when the plane cuts the cone at an angle that is not perpendicular but does not cross the base.
- Parabolas: Created when the plane is parallel to a generator of the cone.
- Hyperbolas: Result from the plane cutting both halves of the double cone.
Standard Form of a Circle
The standard form of a circle's equation is recognizable and distinct. It follows the pattern \( (x-h)^2 + (y-k)^2 = r^2 \), where \( r \) is the radius of the circle, and \( (h, k) \) are the coordinates of the circle's center. The values of \( h \) and \( k \) can be found by examining the equation when it is written as a completed square. In the initial problem statement, after rearranging and completing the square, we found a form similar to the standard circle equation:
The presence of both \( x \) and \( y \) terms squared and on the same side of the equation, as well as the presence of a constant term on the opposite side, indicates that we're dealing with a circle. The conversion to standard form not only aids in identification but also provides immediate insights into the circle's properties, such as its radius and the location of its center.
The presence of both \( x \) and \( y \) terms squared and on the same side of the equation, as well as the presence of a constant term on the opposite side, indicates that we're dealing with a circle. The conversion to standard form not only aids in identification but also provides immediate insights into the circle's properties, such as its radius and the location of its center.
Other exercises in this chapter
Problem 61
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