Problem 7 - Algebra and Trigonometry

Question

In Exercises \(1-18,\) graph each ellipse and locate the foci. $$ \frac{x^{2}}{49}+\frac{y^{2}}{81}=1 $$

Step-by-Step Solution

Answer

The ellipse has its center at the origin \((0,0)\), the lengths of the major and minor axes are 18 and 14 units respectively. The foci of the ellipse are located at \((0, \pm 4\sqrt{2})\).

Step 1: Determine the Center

The given ellipse equation is in standard form \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), where \(a\) and \(b\) are the semi-major and semi-minor axes respectively, if \(b>a\), otherwise \(a\) and \(b\) are the semi-minor and semi-major axes respectively. And the center of the ellipse is at the origin \((0,0)\).

Step 2: Determine the Lengths of the Axes

The lengths of the semi-major axis and semi-minor axis can be extracted directly from the equation as \(a=\sqrt{49}=7\) and \(b=\sqrt{81}=9\). So, the lengths of the major and minor axes are \(2b = 2*9 = 18\) and \(2a = 2*7 = 14\) respectively.

Step 3: Determine the Foci

The foci are located at a distance \(c\) away from the center along the major axis, where \(c = \sqrt{b^2 - a^2} = \sqrt{81 - 49} = \sqrt{32} = 4\sqrt{2}\). Therefore the foci are at \((0, \pm 4\sqrt{2})\).

Step 4: Graph the Ellipse

Plot the center at \((0,0)\). Draw the major axis of length 18 units vertically centered at the origin. Draw the minor axis of length 14 units horizontally, also centered at the origin. Finally, mark the positions of the foci along the vertical axis about \(4\sqrt{2}\) units above and below the origin.

Key Concepts

Standard Form of Ellipse Equation

Understanding the standard form of an ellipse equation is key to success in graphing these curves. An ellipse, one of the conic sections, can be represented mathematically in its standard form as \[\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1\]\ for a horizontal major axis, or \[\frac{x^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} = 1\]\ for a vertical major axis, depending on whether the ellipse is wider across the x-axis or the y-axis, respectively. Here, \(a\) and \(b\) represent the lengths of the semi-major and semi-minor axes, respectively, when \(a > b\). These constants denote the 'radii' in the directions of the axes, essential in plotting the shape of the ellipse.

When students tackle these problems, they should remember that the equation’s form will directly inform them about the ellipse’s orientation, and which axis is the major one—knowledge which is crucial for accurate graphing.

Semi-Major and Semi-Minor Axes

The semi-major and semi-minor axes are central to the shape and size of an ellipse. In the given equation \[\frac{x^{2}}{49} + \frac{y^{2}}{81} = 1\]\, the semi-major axis, \(b\), is the longer one, corresponding to the larger denominator under \(y^2\), and the semi-minor axis, \(a\), is the shorter, linked to the smaller denominator under \(x^2\). In this example, \(b\) would be \(\sqrt{81}=9\) and \(a\) would be \(\sqrt{49}=7\).

The lengths of the actual axes are twice the values of \(a\) and \(b\) because they span from the negative to the positive side of each axis. Therefore, understanding these parameters is central to sketching an accurate representation of an ellipse. We encourage students to visualize the axes as guidelines around which the ellipse stretches out.

Locating the Foci of an Ellipse

In the adventure of graphing an ellipse, pinpointing the foci is like discovering hidden treasure—they’re special points that help define the precise curvature of the ellipse. The foci of an ellipse are found along its major axis, equidistant from the center. For the exercise \[\frac{x^{2}}{49} + \frac{y^{2}}{81} = 1\]\, we calculate the distance of the foci from the center using the equation \(c = \sqrt{b^2 - a^2}\). Here, we have \(c = \sqrt{81 - 49} = 4\sqrt{2}\).

Armed with this knowledge, students can place the foci at \((0, \pm 4\sqrt{2})\) on the vertical axis, since we've already established that this ellipse stretches further along the y-axis. By locating the foci accurately, students ensure the ellipse's shape is precise and true to the characteristics dictated by its equation.