Chapter 12: Problem 5
Find the vertices and foci of the ellipse. Sketch its graph, showing the foci. $$ 5 x^{2}+2 y^{2}=10 $$
Short Answer
Expert verified
Vertices: (0, ±2.24); Foci: (0, ±1.73)
Step by step solution
01
Simplify the equation
Start by dividing both sides by 10 to get the equation in a standard form: \[ \frac{5x^2}{10} + \frac{2y^2}{10} = \frac{10}{10} \] which simplifies to \[ \frac{x^2}{2} + \frac{y^2}{5} = 1 \]. This is now the equation of the ellipse in standard form.
02
Identify the parameters of the ellipse
The equation \( \frac{x^2}{2} + \frac{y^2}{5} = 1 \) can be compared with the standard form of an ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). Here, \( a^2 = 2 \) and \( b^2 = 5 \). Since \( b^2 > a^2 \), this is a vertical ellipse.
03
Calculate semi-axes lengths
Calculate \( a \) and \( b \) from \( a^2 \) and \( b^2 \): \[ a = \sqrt{2} \approx 1.41 \] \[ b = \sqrt{5} \approx 2.24 \] These are the lengths of the semi-major and semi-minor axes, respectively.
04
Find the vertices of the ellipse
Since this is a vertical ellipse, the vertices will be along the y-axis: The vertices are at points \( (0, \pm b) = (0, \pm 2.24) \).
05
Calculate the distance of foci from the center
Use the formula for the distance to the foci: \( c^2 = b^2 - a^2 \). Substitute the values: \[ c^2 = 5 - 2 = 3 \] \[ c = \sqrt{3} \approx 1.73 \].
06
Determine the foci of the ellipse
For a vertical ellipse, the foci are along the y-axis as well. The foci are at \( (0, \pm c) = (0, \pm 1.73) \).
07
Sketch the graph
Draw the major and minor axes. Place the center of the ellipse at the origin (0,0).
Plot the vertices at (0, ±2.24) and the foci at (0, ±1.73). Draw an oval shape around these points to complete the ellipse.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vertices of Ellipse
In the geometry of ellipses, **vertices** are the furthest points along the main axis of the ellipse. For the given equation of an ellipse \[ \frac{x^2}{2} + \frac{y^2}{5} = 1 \],we see that the ellipse is vertical based on the greater denominator being under \(y^2\). This vertical orientation tells us that the **vertices** are aligned along the y-axis.
- The vertices of such a vertical ellipse can be located at \((0, \pm b)\), where \(b\) denotes the length of the semi-major axis.
- Given \(b = \sqrt{5} \approx 2.24\), the vertices are \((0, -2.24)\) and \((0, 2.24)\).
Foci of Ellipse
The **foci** (plural of focus) of an ellipse are two special points on the major axis. The sum of the distances from these points to any point on the ellipse is a constant. For ellipse calculations, the distance from the center to each focus, denoted by \(c\), plays a crucial role.Calculate \(c\) using the formula:\[ c^2 = b^2 - a^2 \],where \(b^2\) and \(a^2\) are the squares of the lengths of the semi-major and semi-minor axes, respectively. For this ellipse, the values are \(b^2 = 5\) and \(a^2 = 2\):
- \(c^2 = 5 - 2 = 3\)
- \(c = \sqrt{3} \approx 1.73\)
Standard Form of Ellipse
The **standard form** of an ellipse is essential as it reveals many properties of the ellipse at a glance. This form is represented mathematically as:\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \] or \[ \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \], depending on whether the ellipse is horizontal or vertical.For the given example, simplifying the initial equation led us to the standard form:\[ \frac{x^2}{2} + \frac{y^2}{5} = 1 \].
- Here, \(a^2 = 2\) and \(b^2 = 5\).
- Since \(b^2 > a^2\), it confirms a vertical ellipse with the major axis along the y-axis.
Semi-Major Axis
For any ellipse, the **semi-major axis** is the longer radius extending from the center of the ellipse to the edge, along the longest line that passes through the ellipse. It plays a vital role in defining the ellipse's size and orientation.In our problem:
- Since \(b^2 = 5 > 2 = a^2\), the semi-major axis is associated with the \(y\) term.
- Thus, the length of the semi-major axis is \(b = \sqrt{5} \approx 2.24\).
Semi-Minor Axis
The **semi-minor axis** is defined as the shortest radius from the ellipse's center to its perimeter. It is shorter than the semi-major axis and, combined with the semi-major axis, defines the ellipse's "width."Following the analysis of our equation:
- Given \(a^2 = 2 < b^2 = 5\), the semi-minor axis corresponds to the \(x\) direction.
- The length of the semi-minor axis is \(a = \sqrt{2} \approx 1.41\).