Problem 10 - Algebra for College Students

Question

Find the vertices, the endpoints of the minor axis, and the foci of the given ellipse, and sketch its graph. See answer section. $$ 8 x^{2}+y^{2}=16 $$

Step-by-Step Solution

Answer

Vertices: (0, 4) and (0, -4); Minor axis endpoints: (\(\sqrt{2}, 0\)) and (\(-\sqrt{2}, 0\)); Foci: (0, \(\sqrt{14}\)) and (0, \(-\sqrt{14}\)).

Step 1: Convert the Equation to Standard Form

The given equation of the ellipse is \(8x^2 + y^2 = 16\). Divide every term by 16 to begin simplifying it. This gives \(\frac{x^2}{2} + \frac{y^2}{16} = 1\). Now the equation resembles the standard form of an ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), where \(a^2 = 2\) and \(b^2 = 16\).

Step 2: Identify the Orientation and Axes Lengths

Since \(b^2 > a^2\), the major axis is vertical. Calculate \(a = \sqrt{2} = \sqrt{2}\) and \(b = \sqrt{16} = 4\). The center of the ellipse is at the origin (0,0) because there are no terms of the form \(hx\) or \(ky\).

Step 3: Find the Vertices

For an ellipse with a vertical major axis, the vertices are located at \((h, k \pm b)\). Since the center is \((0,0)\), the vertices are \((0,0 \pm 4)\), resulting in the points (0, 4) and (0, -4).

Step 4: Find the Endpoints of the Minor Axis

For the endpoints of the minor axis for a vertical ellipse, use \((h \pm a, k)\). This means the endpoints are \((0 \pm \sqrt{2}, 0)\), which are the points (\(\sqrt{2}, 0\)) and (\(-\sqrt{2}, 0\)).

Step 5: Calculate the Foci

The foci of a vertical ellipse can be found using \(c = \sqrt{b^2 - a^2}\). Calculate \(c = \sqrt{16 - 2} = \sqrt{14}\). The foci are located at \((h, k \pm c)\), which translates to \((0, 0 \pm \sqrt{14})\) or (0, \(\sqrt{14}\)) and (0, \(-\sqrt{14}\)).

Key Concepts

Ellipse Equation

An ellipse is a geometric shape that looks like an elongated circle. The equation of an ellipse forms the foundation for understanding its geometry. To work with an ellipse, we often use its standard form equation, which is \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \] when the center is at the origin (0,0). Here, \(a\) and \(b\) are the semi-major and semi-minor axes lengths. For our specific problem, we adjusted the given equation, \(8x^2 + y^2 = 16\), to \[ \frac{x^2}{2} + \frac{y^2}{16} = 1, \] resembling the standard form. In this equation, the terms \(a^2 = 2\) and \(b^2 = 16\) provide crucial information about the ellipse's orientation and dimensions.

Vertices of an Ellipse

The vertices are the 'tip points' of the ellipse along its longest dimension, known as the major axis. To find these, we discern whether the ellipse is vertically or horizontally oriented by comparing \(a^2\) and \(b^2\).

• If \(b^2 > a^2\), the ellipse is oriented vertically, and the major axis runs parallel to the y-axis.
• The vertices would then be at \((h, k \pm b)\), where (h, k) is the center of the ellipse.

In our exercise:

• Since the center is at (0,0), and \(b = 4\), the vertices are (0, 4) and (0, -4).

These points signify the farthest extent of the ellipse along the y-axis.

Foci of an Ellipse

The foci are unique points inside the ellipse, located along the major axis, possessing a special geometric property: the sum of distances from any point on the ellipse to the two foci is constant. Determining the foci positions involves calculating \(c\) using the formula \[ c = \sqrt{b^2 - a^2} \] for a vertically oriented ellipse.

• In this problem, \(b^2 = 16\) and \(a^2 = 2\).
• Thus, \(c = \sqrt{16 - 2} = \sqrt{14}\).

The foci are positioned at \((h, k \pm c)\). With a center of (0,0), our foci become (0, \(\sqrt{14}\)) and (0, \(-\sqrt{14}\)). This configuration further underscores the vertical arrangement of the ellipse.

Minor Axis of an Ellipse

The minor axis of an ellipse is its shortest diameter. It is perpendicular to the major axis and intersects it at the center of the ellipse.

• For our ellipse, the minor axis is horizontal because the major axis is vertical.
• The minor axis endpoints are calculated using \((h \pm a, k)\), where \(a = \sqrt{2}\).

In the given solution:

• The endpoints of the minor axis are at \((\sqrt{2}, 0)\) and \((-\sqrt{2}, 0)\).

These points represent the extent of the ellipse along the x-axis, giving it that classic oval shape.