Problem 17 - College Algebra

Question

In Exercises \(1-18,\) graph each ellipse and locate the foci. $$7 x^{2}=35-5 y^{2}$$

Step-by-Step Solution

Answer

The graph of the ellipse will be centred at the origin with a horizontal extent of \(2\sqrt{5}\) and a vertical extent of \(2\sqrt{7}\). The foci are located at \((0, \pm \sqrt{2})\) on the vertical axis.

Step 1: Rearrange the equation in standard form

Rearranging the given equation \(7x^2 = 35 - 5y^2\), we want to get everything on one side so that it's equal to 1. Rearranging gives \(\frac{x^2}{5} - \frac{y^2}{7} = 1\).

Step 2: Identify the lengths of the semi-major and semi-minor axes

Here, \(a^2 = 5\) and \(b^2 = 7\), so \(a = \sqrt{5}\) and \(b = \sqrt{7}\). Since \(b^2 > a^2\), the major axis is vertical.

Step 3: Calculate the foci

The foci are given by \((h, k \pm c)\), where \(c = \sqrt{b^2 - a^2}\). So, \(c = \sqrt{7 - 5} = \sqrt{2}\). So, the foci of the ellipse are \((0, \pm \sqrt{2})\).

Step 4: Graph the ellipse

Finally, to graph the ellipse, draw the major axis vertically centred at the origin with length \(2b = 2\sqrt{7}\) and the minor axis horizontally also centred at the origin with length \(2a = 2\sqrt{5}\). The foci will be located at \((0, \pm \sqrt{2})\) on the vertical axis.

Key Concepts

Standard Form of Ellipse

Understanding the standard form of an ellipse is foundational to graphing and analyzing its properties. An ellipse's equation in standard form is given by \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) for a horizontal ellipse, or \(\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1\) for a vertical ellipse, where \(a\) and \(b\) are the lengths of the semi-major and semi-minor axes, respectively.

In the exercise \(7x^2 = 35 - 5y^2\), the equation was first rearranged into \(\frac{x^2}{5} - \frac{y^2}{7} = 1\) which isn’t quite the standard form yet due to the negative sign. To correct this, we divide everything by \( -1\), which yields \(\frac{y^2}{7} - \frac{x^2}{5} = -1\); a form closer to that of a hyperbola. To obtain the standard form of an ellipse, we need an equation equal to \(1\) with positive coefficients. Hence, the correct manipulation would involve moving terms and flipping fractions to get \(\frac{x^2}{5} + \frac{y^2}{35/5} = 1\), revealing that \(a^2 = 5\) and \(b^2 = 7\), representing a vertical ellipse.

Semi-Major and Semi-Minor Axes

The semi-major and semi-minor axes are imperative for describing the shape and dimensions of an ellipse. For a vertical ellipse, the semi-major axis (\(a\)) is the longer one and runs vertically, while the semi-minor axis (\(b\)) is shorter and runs horizontally. Conversely, for a horizontal ellipse, these definitions are switched.

From the rearranged equation, \(a^2 = 5\) and \(b^2 = 7\) indicate that \(a = \sqrt{5}\) and \(b = \sqrt{7}\). Importantly, since \(b^2 > a^2\), our ellipse is vertically oriented with the major axis longer than the minor axis, which is a key point to consider for accurate graphing.

Ellipse Foci Calculation

The ellipse's foci are two fixed points located along the major axis, equidistant from the center, and are crucial in the ellipse's definition. To find the foci, one must calculate \(c\) as \(c = \sqrt{\lvert b^2 - a^2 \rvert}\) for a vertical ellipse, where \(c\) represents the distance from the center to each focus.

In our example, since \(b^2 > a^2\), we determine \(c\) using \(c = \sqrt{b^2 - a^2} = \sqrt{7 - 5} = \sqrt{2}\). Therefore, the foci of our ellipse are located at \(0, \pm \sqrt{2}\), which aligns with the major axis being vertical.

Ellipse Graphing Steps

Graphing an ellipse involves clear steps, starting with its standard form. Begin by identifying the center which, in our case, is at the origin \(0,0\) since no \(h\) or \(k\) values are present in the standard form. Draw the major axis along the appropriate orientation; in this instance, vertically, and plot points at \(\pm a\) distances from the center along this axis.

Next, draw the minor axis perpendicular to the major axis, here, horizontally, and mark points at \(\pm b\) from the center. Then, sketch the ellipse, ensuring it passes through these four points. Finally, accurately place the foci on the major axis at the calculated \(c\) distances from the center, completing the graph. Using the given equation, we would draw the major axis with a length of \(2\sqrt{7}\) and the minor axis with a length of \(2\sqrt{5}\), placing the foci at \(0, \pm \sqrt{2}\).