Problem 18 - Algebra 2

Question

Find the coordinates of the center and foci and the lengths of the major and minor axes for the ellipse with the given equation. Then graph the ellipse. \(\frac{y^{2}}{10}+\frac{x^{2}}{5}=1\)

Step-by-Step Solution

Answer

Center: (0, 0), Foci: (0, ±√5), Lengths: Major 2√10, Minor 2√5.

Step 1: Identify the Equation Form

The given equation is \( \frac{y^2}{10} + \frac{x^2}{5} = 1 \). This resembles the standard form of an ellipse equation \( \frac{y^2}{a^2} + \frac{x^2}{b^2} = 1 \), where if \( a > b \), the major axis is along the y-axis.

Step 2: Determine the Values of a and b

In this case, we compare the given equation to the standard form and identify \( a^2 = 10 \) and \( b^2 = 5 \). Thus, \( a = \sqrt{10} \) and \( b = \sqrt{5} \). Since \( a > b \), the major axis is along the y-axis.

Step 3: Find the Center of the Ellipse

The standard form of the ellipse centered at the origin is \( \frac{(y-k)^2}{a^2} + \frac{(x-h)^2}{b^2} = 1 \). Since there are no \( (x-h) \) or \( (y-k) \) terms, the center \((h, k)\) is \( (0, 0) \).

Step 4: Calculate the Lengths of the Axes

The major axis length is \( 2a = 2\sqrt{10} \) and the minor axis length is \( 2b = 2\sqrt{5} \).

Step 5: Determine the Foci

The distance from the center to each focus is \( c \), found using \( c^2 = a^2 - b^2 \). Thus, \( c^2 = 10 - 5 = 5 \), so \( c = \sqrt{5} \). The foci are located at \((0, \pm \sqrt{5})\).

Step 6: Graph the Ellipse

With the above information, the ellipse can be graphed with the center at \((0,0)\), vertices at \((0, \pm \sqrt{10})\), co-vertices at \((\pm \sqrt{5}, 0)\), and foci at \((0, \pm \sqrt{5})\).

Key Concepts

Center

The center of an ellipse is the midpoint between its vertices and co-vertices. In mathematical terms, for an ellipse in the form \[\frac{(y-k)^2}{a^2} + \frac{(x-h)^2}{b^2} = 1\]the center is \((h, k)\).In our exercise, the ellipse equation is \[\frac{y^2}{10} + \frac{x^2}{5} = 1\],which is already centered at the origin \((0, 0)\).This means there are no shifts along the x-axis or y-axis, confirming that the center of this ellipse is precisely at the origin. This central point is the reference for locating other significant features of the ellipse, such as the foci and vertices.

In summary:

• Equation: Standard form \(\frac{y^2}{10} + \frac{x^2}{5} = 1\)
• Center coordinates: \((0, 0)\)

Foci

The foci of an ellipse are two distinct points along the major axis, symmetrically positioned from the center. The foci are vital in defining the shape of the ellipse. To calculate the distance to the foci, we use the equation:\[c^2 = a^2 - b^2\]where \(a\) is the semi-major axis and \(b\) is the semi-minor axis.For the given ellipse \(\frac{y^2}{10} + \frac{x^2}{5} = 1\),we've previously determined that \(a^2 = 10\)and \(b^2 = 5\).Thus:

• \(c^2 = 10 - 5 = 5\)
• \(c = \sqrt{5}\)

The foci are therefore located at \((0, \pm \sqrt{5})\).Given that the major axis is along the y-axis due to \(a > b\),these foci are vertically aligned above and below the center by \(\sqrt{5}\) units.

Major Axis

The major axis of an ellipse is the longest diameter and passes through the center. It defines the direction in which the ellipse is elongated. In standard form, the major axis is determined by the larger of either \(a\) or \(b\).For the ellipse \(\frac{y^2}{10} + \frac{x^2}{5} = 1\),we found \(a^2 = 10\) and \(b^2 = 5\),thus making \(a = \sqrt{10}\) greater than \(b = \sqrt{5}\).This is why the major axis is along the y-axis. The length of the major axis is calculated as \(2a\),so in our case:

• Major axis length: \(2 \times \sqrt{10} = 2\sqrt{10}\)

This axis stretches vertically, meaning the ellipse extends more in the y-direction, resulting in vertices at \((0, \pm \sqrt{10})\).
This provides a clear guideline for the ellipse's shape.

Minor Axis

The minor axis of an ellipse is the shortest diameter, perpendicular to the major axis, and similarly passes through the center of the ellipse.For the given ellipse equation: \(\frac{y^2}{10} + \frac{x^2}{5} = 1\),we determined \(b^2 = 5\),which makes \(b = \sqrt{5}\).The minor axis lies along the x-axis because it is associated with the smaller denominator in the equation.The length of the minor axis is calculated as \(2b\):

• Minor axis length: \(2 \times \sqrt{5} = 2\sqrt{5}\)

This axis provides the bounds for the ellipse in the x-direction, resulting in co-vertices at \((\pm \sqrt{5}, 0)\).These locations mark the points where the ellipse reaches its maximum extent along the minor axis, giving the ellipse its overall oval shape.