Chapter 9: Problem 28
Write the equation of the given ellipse in standard form. . \(x^{2}+y^{2}-4 x-4 y+4=0\)
Short Answer
Expert verified
The equation in standard form is \(\frac{(x-2)^2}{4} + \frac{(y-2)^2}{4} = 1\).
Step by step solution
01
Rearrange the Equation
Start with the given equation: \[x^2 + y^2 - 4x - 4y + 4 = 0\]Move the constant to the other side:\[x^2 - 4x + y^2 - 4y = -4\]
02
Complete the Square for x-terms
Take the x-terms and complete the square:The expression is \(x^2 - 4x\).Take half of -4 (\(-\frac{4}{2} = -2\)), and square it (\((-2)^2 = 4\)): Add and subtract 4 inside the equation:\[(x^2 - 4x + 4)\]This simplifies to \((x-2)^2\).
03
Complete the Square for y-terms
Take the y-terms and complete the square:The expression is \(y^2 - 4y\). Take half of -4 (\(-\frac{4}{2} = -2\)), and square it (\((-2)^2 = 4\)): Add and subtract 4 inside the equation:\[(y^2 - 4y + 4)\]This simplifies to \((y-2)^2\).
04
Write the Completed Squares into the Equation
Insert the completed squares back into the equation:\[(x-2)^2 + (y-2)^2 = -4 + 4 + 4\]Simplify the expression:\[(x-2)^2 + (y-2)^2 = 4\]
05
Identify the Standard Form of the Ellipse
The equation \((x-2)^2 + (y-2)^2 = 4\) is already in the standard form of an ellipse centered at \((h, k) = (2, 2)\) with \(a^2 = 4\) and \(b^2 = 4\). The equation \(\frac{(x-2)^2}{4} + \frac{(y-2)^2}{4} = 1\) indicates it is a circle, a special case of an ellipse.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Completing the Square
Completing the square is an essential method used to simplify quadratic expressions and put them in a form that is easy to analyze. In the context of converting a general conic equation to its standard form, completing the square is crucial. Here's how it works:
- Identify the quadratic and linear terms for each variable in the equation.
- For instance, when working with the quadratic expression \( x^2 - 4x \), observe the coefficient of \( x \), which is -4.
- Take half of this coefficient: \(-\frac{4}{2} = -2\).
- Square this result to complete the square: \((-2)^2 = 4\).
- Add and subtract this new term (4) inside the expression to rewrite it as a perfect square trinomial. This gives us \((x-2)^2\).
Ellipse Equation
An ellipse is a type of conic section that brings together a family of curves defined within a plane. When an ellipse is centered at a point other than the origin, its equation has a unique form. Transforming an ellipse equation to its standard form allows us to understand its properties more clearly.
- The standard form of an ellipse equation with a center at \((h, k)\) is presented as \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \).
- Here, \( (h, k) \) is the center of the ellipse.
- The values \( a^2 \) and \( b^2 \) represent squares of semi-major and semi-minor axes respectively. If \( a = b \), then the ellipse becomes a circle.
Geometry of Conics
Conic sections like ellipses represent the intersection of a plane and a double-napped cone. The study of these curves, which also includes hyperbolas, parabolas, and circles, is known as the geometry of conics. Understanding conics is fundamental to both geometry and algebra.
- Each conic section, including ellipses, can be characterized by its unique standard equation and geometric properties.
- The orientation and position are determined by parameters within the equation, such as the center, axes, and relative size.
- Ellipses, in particular, have special attributes, like having two focal points and the sum of the distances from any point on the ellipse to the foci being constant.