Problem 2 - Algebra and Trigonometry

Question

The graph of the equation \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) with \(a>b>0\) is an ellipse with vertices (_, _) and (_, _) and foci \(( \pm c, 0)\) where \(c=\) _________ So the graph of \(\frac{x^{2}}{5^{2}}+\frac{y^{2}}{4^{2}}=1\) is an ellipse with vertices (_, _) and (_, _) and foci (_, _) and (_, _)

Step-by-Step Solution

Answer

Vertices: (5, 0) and (-5, 0); Foci: (3, 0) and (-3, 0).

Step 1: Identify the Standard Ellipse Equation

The general form of the ellipse equation is given by \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) where \( a \) is the semi-major axis and \( b \) is the semi-minor axis. Here, the major axis is along the x-axis because \( a > b \).

Step 2: Extract 'a' and 'b' from the Given Equation

The given ellipse equation is \( \frac{x^2}{5^2} + \frac{y^2}{4^2} = 1 \). Thus, \( a = 5 \) and \( b = 4 \).

Step 3: Calculate the Foci of the Ellipse

The foci \((\pm c, 0)\) are calculated using \( c = \sqrt{a^2 - b^2} \). Substituting \( a = 5 \) and \( b = 4 \), we find \( c = \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \sqrt{9} = 3 \). So, the foci are \((\pm 3, 0)\), which means (3, 0) and (-3, 0).

Step 4: Determine the Vertices of the Ellipse

For an ellipse centered at (0, 0) along the x-axis, the vertices are at (\( \pm a, 0 \)). Since \( a = 5 \), the vertices are (5, 0) and (-5, 0).

Key Concepts

Standard Equation of an Ellipse

An ellipse is a smooth, closed curve that is defined by its standard equation. The equation looks like this: \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). In this form, \(a\) and \(b\) are crucial values that help us describe the shape and position of the ellipse in a coordinate plane.
The terms \(x^2/a^2\) and \(y^2/b^2\) represent the coordinates \(x\) and \(y\) being divided or stretched along their respective axes. Here, \(a\) represents the semi-major axis, and \(b\) stands for the semi-minor axis.

• If \(a > b\), then the ellipse is elongated along the x-axis.
• If \(b > a\), the ellipse is stretched along the y-axis.

This construction is important because it determines how elongated or circular the ellipse appears. In our example, both \(a\) and \(b\) are real numbers, with \(a = 5\) and \(b = 4\). So, the ellipse is longer in the x-direction.

Vertices of an Ellipse

The vertices of an ellipse are specific points where the ellipse reaches its widest span along the major axis. These are important because they help in understanding the overall size and orientation of the ellipse in the plane.
For an ellipse centered at the origin along the x-axis, like our example, the vertices are located at coordinates \((\pm a, 0)\). This is because the major axis is along the x-axis, and \(a\) tells us how far the ellipse stretches in that direction.

• In the given ellipse equation, we've identified \(a = 5\).
• Thus, the vertices are at \((5, 0)\) and \((-5, 0)\).

These vertices are essential landmarks on the ellipse's geometry, defining how wide it is across its primary axis of symmetry.

Foci of an Ellipse

The foci (singular: focus) of an ellipse are two key points inside the ellipse that help characterize its shape. The role of the foci is special in the ellipse's definition: the sum of the distances from any point on the ellipse to the foci is constant.
To find the foci, we use the formula \(c = \sqrt{a^2 - b^2}\) where \(c\) is the distance from the center of the ellipse to each focus.
Using our example values:

• \(a = 5\), \(b = 4\)
• Calculate \(c = \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \sqrt{9} = 3\)

Thus, in our ellipse, these points are positioned at \((\pm 3, 0)\), which translate to points \((3, 0)\) and \((-3, 0)\). These foci provide insight into how much the ellipse deviates from being a perfect circle, with more elongated ellipses having foci further apart.