Problem 7 - 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. $$ 2 x^{2}+5 y^{2}=50 $$

Step-by-Step Solution

Answer

Vertices: (-5,0), (5,0); Endpoints Minor Axis: (0,-√10), (0,√10); Foci: (-√15,0), (√15,0).

Step 1: Standard Form of the Ellipse

First, we'll rewrite the given equation \(2x^2 + 5y^2 = 50\) in standard form. Divide each term by 50 to set the equation equal to 1:\[ \frac{2x^2}{50} + \frac{5y^2}{50} = 1 \]Simplify:\[ \frac{x^2}{25} + \frac{y^2}{10} = 1 \]This equation shows the ellipse centered at the origin \((0,0)\) with the semi-major axis along the x-axis.

Step 2: Identify Semi-Major and Semi-Minor Axes

Compare the standard form to the general ellipse form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\).Here, \(a^2 = 25\) and \(b^2 = 10\), so \(a = 5\) and \(b = \sqrt{10}\).Thus, the semi-major axis is 5 units along the x-axis, and the semi-minor axis is \(\sqrt{10}\) units along the y-axis.

Step 3: Find the Vertices

The vertices are at the ends of the major axis. Given that \(a = 5\), the vertices along the x-axis are:\((-5,0)\) and \((5,0)\).

Step 4: Find the Endpoints of the Minor Axis

The endpoints of the minor axis are at y coordinates \(\pm b\), where \(b = \sqrt{10}\). Thus the endpoints along the y-axis are:\((0, -\sqrt{10})\) and \((0, \sqrt{10})\).

Step 5: Determine the Foci

The foci are found using the formula \(c^2 = a^2 - b^2\).\[ c^2 = 25 - 10 = 15 \]Thus, \(c = \sqrt{15}\), so the foci are \((\pm \sqrt{15}, 0)\) along the x-axis.

Step 6: Graph the Ellipse

With the vertices \((-5,0)\) and \((5,0)\), the minor axis endpoints \((0, -\sqrt{10})\) and \((0, \sqrt{10})\), and the foci \((\pm \sqrt{15}, 0)\), sketch the ellipse centered at the origin, elongating along the x-axis due to larger semi-major axis length.

Key Concepts

Vertices

In an ellipse, the vertices are the points where the curve meets the ends of its major axis. To find the vertices of an ellipse given in its equation form, like \( \frac{x^2}{25} + \frac{y^2}{10} = 1 \), we first need to understand which term corresponds to the major axis. Since \( a^2 = 25 \) is larger than \( b^2 = 10 \), the major axis is along the x-axis. The vertices will be at \( (\pm a, 0) \) for a horizontally stretched ellipse. So, for our equation, the value of \( a \) is 5. Therefore, the vertices are located at:

• \( (-5, 0) \)
• \( (5, 0) \)

This means the vertices are 10 units apart, centering the ellipse along the x-axis.

Minor Axis

The minor axis of an ellipse is perpendicular to the major axis and shorter in length. For our ellipse equation \( \frac{x^2}{25} + \frac{y^2}{10} = 1 \), since \( b^2 = 10 \), the minor axis is along the y-axis.To find the endpoints of the minor axis, we use \( b \), the distance from the center to the endpoint:

• \( b = \sqrt{10} \)

Thus, the endpoints of the minor axis will be:

• \( (0, \sqrt{10}) \)
• \( (0, -\sqrt{10}) \)

These endpoints indicate that the minor axis of the ellipse is symmetric around the y-axis at the origin, forming a vertical line segment through the ellipse center.

Foci

Foci are two special points located inside the ellipse along the major axis. They play an important role in the reflective property of an ellipse, as the sum of the distances from the foci to any point on the ellipse is constant. For the equation \( \frac{x^2}{25} + \frac{y^2}{10} = 1 \), we use the relationship \( c^2 = a^2 - b^2 \) to find the distance of each focus from the center along the x-axis.Since \( a^2 = 25 \) and \( b^2 = 10 \), calculate \( c \) as follows:

• \( c^2 = 25 - 10 = 15 \)
• \( c = \sqrt{15} \)

This gives us the foci positions:

• \( (\sqrt{15}, 0) \)
• \( (-\sqrt{15}, 0) \)

These points lie within the body of the ellipse, closer to the center than the vertices.

Standard Form of an Ellipse

Understanding the standard form of an ellipse is essential to accurately graph and identify its dimensions and orientation. An ellipse can have its equation written in the form:\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]If an ellipse equation like \( 2x^2 + 5y^2 = 50 \) is given, we convert it to standard form by dividing through by 50. This sets our equation to:\[ \frac{x^2}{25} + \frac{y^2}{10} = 1 \]In this form:- **\(a^2\)**: Denominator of the \(x^2\) term, determines major axis length- **\(b^2\)**: Denominator of the \(y^2\) term, determines minor axis length- If \(a^2 > b^2\), the ellipse is stretched along the x-axisKnowing these allows us to quickly find dimensions, such as vertex and foci placement, and determine which axis is the major or minor, making sketching and analyzing the ellipse much easier.