Problem 31
Question
The center of an ellipse or hyperbola is the point of intersection of its axes of symmetry. Each state whether the graph of the given Cartesian equation is an ellipse or hyperbola. Determine its standard form and center. \(x^{2}+2 x+4 v^{2}+24 y=12\)
Step-by-Step Solution
Verified Answer
It's an ellipse with a center at (-1, -3).
1Step 1: Write Down the Original Equation
The given equation is \( x^2 + 2x + 4y^2 + 24y = 12 \). We're going to transform it to identify whether it represents an ellipse or hyperbola.
2Step 2: Rearrange the Equation
Separate the \( x \)-terms and the \( y \)-terms: \[ (x^2 + 2x) + (4y^2 + 24y) = 12. \]
3Step 3: Complete the Square for x-terms
Take the \( x \)-terms, \( x^2 + 2x \), and complete the square: Calculate \( \left(\frac{2}{2}\right)^2 = 1 \). Add and subtract this inside the parenthesis to get:\( (x^2 + 2x + 1 - 1) = (x+1)^2 - 1 \).
4Step 4: Complete the Square for y-terms
Take the \( y \)-terms, \( 4y^2 + 24y \), and factor out 4: \( 4(y^2 + 6y) \).Complete the square: Calculate \( \left(\frac{6}{2}\right)^2 = 9 \). Add and subtract within the parenthesis:\( 4((y^2+6y+9-9) = 4((y+3)^2 - 9) \).
5Step 5: Simplify the Equation
Substitute back the completed squares back into the equation:\[ (x+1)^2 - 1 + 4((y+3)^2 - 9) = 12. \]Simplify:\[ (x+1)^2 + 4(y+3)^2 - 1 - 36 = 12, \]\[ (x+1)^2 + 4(y+3)^2 = 49. \]
6Step 6: Identify the Conic Section and Standard Form
Divide the entire equation by 49:\[ \frac{(x+1)^2}{49} + \frac{4(y+3)^2}{49} = 1. \]Rewrite as: \[ \frac{(x+1)^2}{49} + \frac{(y+3)^2}{ rac{49}{4}} = 1. \]Since this equation takes the form \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), it represents an ellipse.
7Step 7: Determine the Center
The equation \( \frac{(x+1)^2}{49} + \frac{(y+3)^2}{ rac{49}{4}} = 1 \) is in the standard form of an ellipse with center at \((-1,-3)\).
Key Concepts
Conic SectionsCompleting the SquareStandard Form of Ellipse
Conic Sections
Conic sections are the curves obtained by intersecting a cone with a plane. They include several key shapes: circles, ellipses, parabolas, and hyperbolas.
Each shape has distinct characteristics and can be identified by the form of its equation.
- **Circle:** This occurs when the plane cuts perpendicular to the cone's axis. Equations are usually in the form \(x^2 + y^2 = r^2\).- **Ellipse:** Formed when the plane cuts obliquely, creating an oval shape with a specific standard form: \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\). Here, \(a\) and \(b\) are the ellipse's semi-major and semi-minor axes.- **Parabola:** A plane parallel to the slant of the cone results in a parabola. It's observed in equations like \(y^2 = 4ax\).- **Hyperbola:** When the plane cuts both nappes of the cone, a hyperbola is formed, displayed by \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\).In our original problem, identifying the conic section involves recognizing that the transformed equation fits the standard form of an ellipse.
Each shape has distinct characteristics and can be identified by the form of its equation.
- **Circle:** This occurs when the plane cuts perpendicular to the cone's axis. Equations are usually in the form \(x^2 + y^2 = r^2\).- **Ellipse:** Formed when the plane cuts obliquely, creating an oval shape with a specific standard form: \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\). Here, \(a\) and \(b\) are the ellipse's semi-major and semi-minor axes.- **Parabola:** A plane parallel to the slant of the cone results in a parabola. It's observed in equations like \(y^2 = 4ax\).- **Hyperbola:** When the plane cuts both nappes of the cone, a hyperbola is formed, displayed by \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\).In our original problem, identifying the conic section involves recognizing that the transformed equation fits the standard form of an ellipse.
Completing the Square
Completing the square is a technique used in algebra to transform a quadratic equation, making it easier to work with.
This involves turning a quadratic expression into a perfect square trinomial.
Here's a closer look at how we complete the square for different terms:- **For \(x\)-terms:** Start with a quadratic expression like \(x^2 + bx\). To complete the square, we calculate \(\left(\frac{b}{2}\right)^2\). Adding and subtracting this value inside the equation transforms it into \((x + \frac{b}{2})^2 - \left(\frac{b}{2}\right)^2\).- **For \(y\)-terms similar approach:** If the expression is something like \(ky^2 + dy\), factor out the coefficient \(k\), then complete the square for the expression \(y^2 + \frac{d}{k}y\).This method is particularly useful for conic sections since it helps in quickly transforming an equation into a recognizable and useful form, such as the standard form of an ellipse or other conics, allowing us to easily identify important features like the center or vertex.
This involves turning a quadratic expression into a perfect square trinomial.
Here's a closer look at how we complete the square for different terms:- **For \(x\)-terms:** Start with a quadratic expression like \(x^2 + bx\). To complete the square, we calculate \(\left(\frac{b}{2}\right)^2\). Adding and subtracting this value inside the equation transforms it into \((x + \frac{b}{2})^2 - \left(\frac{b}{2}\right)^2\).- **For \(y\)-terms similar approach:** If the expression is something like \(ky^2 + dy\), factor out the coefficient \(k\), then complete the square for the expression \(y^2 + \frac{d}{k}y\).This method is particularly useful for conic sections since it helps in quickly transforming an equation into a recognizable and useful form, such as the standard form of an ellipse or other conics, allowing us to easily identify important features like the center or vertex.
Standard Form of Ellipse
Understanding the standard form of an ellipse is crucial when dealing with conic sections.
This standard form provides a clear view of an ellipse's dimensions and orientation.
The standard form for a horizontal ellipse is:\[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\]Here:
The center of the ellipse was identified as \((-1, -3)\) after completing the square.
This process highlighted the necessity of rearranging and simplifying the equation to match the standard form, providing structured insights into the ellipse's geometry.
This standard form provides a clear view of an ellipse's dimensions and orientation.
The standard form for a horizontal ellipse is:\[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\]Here:
- \((h, k)\) represents the center of the ellipse.
- \(a\) is the semi-major axis, which is the longest radius of the ellipse.
- \(b\) is the semi-minor axis, the shortest radius.
The center of the ellipse was identified as \((-1, -3)\) after completing the square.
This process highlighted the necessity of rearranging and simplifying the equation to match the standard form, providing structured insights into the ellipse's geometry.
Other exercises in this chapter
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