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Magazine Chapter 10: Problem 58
The graph of equation is an ellipse. Determine which distance is longer, the distance between the \(x\)-intercepts or the distance between the \(y\) -intercepts. How much longer? \(x^{2}+4 y^{2}=36\)
Short Answer
Expert verified
The x-intercepts are 6 units longer than the y-intercepts.
Step by step solution
01
Rewrite the Equation in Standard Form
The given equation is \( x^2 + 4y^2 = 36 \). Divide every term by 36 to rewrite it in standard form of an ellipse, \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). We get: \( \frac{x^2}{36} + \frac{y^2}{9} = 1 \). This corresponds to \( a^2 = 36 \) and \( b^2 = 9 \), so \( a = 6 \) and \( b = 3 \).
02
Determine the x-Intercepts
To find the \( x \)-intercepts, set \( y = 0 \) in the standard form of the ellipse: \( \frac{x^2}{36} = 1 \). Solving for \( x \), we get \( x^2 = 36 \), so \( x = \pm 6 \). The \( x \)-intercepts are \( (6, 0) \) and \( (-6, 0) \).
03
Calculate the Distance Between x-Intercepts
The distance between the points \( (6, 0) \) and \( (-6, 0) \) on the x-axis is \( 6 - (-6) = 12 \).
04
Determine the y-Intercepts
To find the \( y \)-intercepts, set \( x = 0 \) in the standard form of the ellipse: \( \frac{y^2}{9} = 1 \). Solving for \( y \), we get \( y^2 = 9 \), so \( y = \pm 3 \). The \( y \)-intercepts are \( (0, 3) \) and \( (0, -3) \).
05
Calculate the Distance Between y-Intercepts
The distance between the points \( (0, 3) \) and \( (0, -3) \) on the y-axis is \( 3 - (-3) = 6 \).
06
Compare the Distances and Find the Difference
The distance between the \( x \)-intercepts is 12, and the distance between the \( y \)-intercepts is 6. The distance between the \( x \)-intercepts is longer by \( 12 - 6 = 6 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Standard Form of an Ellipse
In geometry, the equation of an ellipse can take a specific structure known as the standard form, which simplifies the way we analyze the ellipse's properties. The standard form of an ellipse is expressed as: \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]Here, the values of \( a^2 \) and \( b^2 \) are essential in determining the dimensions of the ellipse:
- \( a \) represents the semi-major axis length if \( a > b \), indicating the longer radius.
- \( b \) represents the semi-minor axis length if \( b < a \), indicating the shorter radius.
- Divide every term by 36 to isolate the ellipse's equation: \( \frac{x^2}{36} + \frac{y^2}{9} = 1 \).
- Identify that \( a^2 = 36 \) and \( b^2 = 9 \), thus \( a = 6 \) and \( b = 3 \).
x-intercepts
Finding the x-intercepts of an ellipse is like locating where the ellipse crosses the x-axis. For this, we set \( y = 0 \) in the standard form equation, which means solving:\[ \frac{x^2}{36} = 1 \]This equation simplifies to:
- \( x^2 = 36 \)
- \( x = \pm 6 \)
y-intercepts
Just as we find the x-intercepts by setting \( y = 0 \), to find the y-intercepts, we set \( x = 0 \) in our ellipse's equation, discovering where it crosses the y-axis. This process involves:\[ \frac{y^2}{9} = 1 \]Upon solving this, it simplifies to:
- \( y^2 = 9 \)
- \( y = \pm 3 \)
Distance Calculation
To fully understand the dimensions of an ellipse, we calculate the distances between the intercepts along each axis. This highlights the width and height of the ellipse:
- **Distance Between x-Intercepts:** - The points \((6, 0)\) and \((-6, 0)\) lie on the x-axis. - The distance formula in one-dimensional terms calculates as: \(|6 - (-6)| = 12\).
- **Distance Between y-Intercepts:** - The points \((0, 3)\) and \((0, -3)\) are on the y-axis. - Similarly, the distance is \(|3 - (-3)| = 6\).
