Problem 14 - Algebra and Trigonometry

Question

In Exercises \(1-18,\) graph each ellipse and locate the foci. $$ 9 x^{2}+4 y^{2}=36 $$

Step-by-Step Solution

Answer

To solve this, we first identified the coefficients 'A', 'B', and 'C'. Then we expressed the equation in standard form. We found the center of the ellipse at the origin (0,0), calculated the foci as (0, ±\sqrt{5}), and the vertices as (±2, 0). Thus, by plotting these points on a graph, we sketched the ellipse.

Step 1: Identify the values of 'A', 'B', and 'C'

The given equation is \(9x^{2} + 4y^{2} = 36\). From this, 'A' can be identified as 9, 'B' as 4, and 'C' as 36.

Step 2: Express the equation in standard form

The standard form of an ellipse's equation is \(\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1\). To express the given equation in standard form, we divide through by 'C', which gives us \(\frac{x^{2}}{4} + \frac{y^{2}}{9} = 1\). Here, \(a^{2} = 4\) and \(b^{2} = 9\).

Step 3: Find the center, foci, and vertices

The center for any general ellipse is at the origin (0,0) in our equation. But to locate the foci, we need to calculate \(c = \sqrt{a^{2} - b^{2}}\), where 'a' represents the semi-major axis and 'b' the semi-minor axis. Since this is a vertical ellipse (because \(b^{2} > a^{2}\)), the foci will be located at (0, ±c). Given that \(a^{2} = 4\) and \(b^{2} = 9\), we find \(c = \sqrt{9 - 4} = \sqrt{5}\). Therefore, the foci are at points (0, ±\sqrt{5}). The vertices, which are the intersection of the ellipse with the x-axis, are at (±2, 0).

Step 4: Graph the ellipse

On the coordinate plane, plot the center of the ellipse at the origin (0,0). Then plot the vertices (±2, 0) along the x-axis and the foci (0, ±\sqrt{5}) along the y-axis. Then sketch the ellipse by drawing a curve through these points. This ellipse opens vertically since \(b^{2}> a^{2}\).

Key Concepts

Conic Sections

Conic sections are the curves obtained by intersecting a plane with a cone. There are four primary types of conic sections: circles, ellipses, parabolas, and hyperbolas. These geometric shapes are foundational in understanding various mathematical concepts and applications, such as planetary orbits or the design of reflective surfaces.
When a plane slices through a cone in such a way that it cuts both nappes but doesn't pass through the apex or base, the resulting shape is an ellipse. This particular conic section is characterized by being a closed curve. Ellipses play a significant role in various fields, such as astronomy, where they describe the orbits of planets and satellites.
Understanding conic sections provides insight into many natural and human-made phenomena, emphasizing the importance of mastering how to work with these shapes.

Standard Form of Ellipse

The standard form of an ellipse's equation is crucial for graphing and understanding its properties. It is expressed as:

• Horizontal ellipse: \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \)
• Vertical ellipse: \( \frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1 \)

Here, \((h, k)\) represents the center of the ellipse. The terms \(a^2\) and \(b^2\) denote the squares of the lengths of the semi-major and semi-minor axes. If \(a^2 > b^2\), the ellipse is oriented horizontally. Conversely, if \(b^2 > a^2\), it is oriented vertically.
Converting the given ellipse equation to its standard form involves algebraically manipulating the terms to match one of these general equations. This makes it simpler to identify key components of the ellipse, such as its axes, center, and orientation.

Foci of Ellipse

The foci are specific points within an ellipse that are crucial in its geometric construction. The sum of the distances from any point on the ellipse to the two foci is constant. This property differentiates an ellipse from other conic sections.
To locate the foci, use the formula \( c = \sqrt{|a^2 - b^2|} \), where \( c \) is the distance from the center to each focus. For a vertical ellipse, the foci are located along the y-axis, and for a horizontal ellipse, they lie along the x-axis.
In our exercise, since the equation is \( \frac{x^2}{4} + \frac{y^2}{9} = 1 \), we have a vertical ellipse where:

• \(a^2 = 4\)
• \(b^2 = 9\)
• \(c = \sqrt{9 - 4} = \sqrt{5}\)
• Foci at \((0, \pm \sqrt{5})\)

Understanding the foci helps comprehend the structure and symmetry this shape possesses internally.

Graphing Ellipses

Graphing an ellipse involves plotting key points and drawing the curve. It starts with determining the ellipse's properties from its standard equation:

• Locate the center, \((h, k)\).
• Identify the vertices, which are points at a distance \(\pm a\) from the center along the semi-major axis.
• Identify the co-vertices, placed at \(\pm b\) along the semi-minor axis.
• Calculate and plot the foci positions based on the calculated value of \(c\).

For our specific example, the ellipse is centered at the origin with vertices \((\pm 2, 0)\) and foci \((0, \pm \sqrt{5})\). The graph of the ellipse will be a smooth curve passing through these points.
The relationship between the semi-major and semi-minor axes determines whether the ellipse is stretched more horizontally or vertically. Understanding these graphical attributes boosts comprehension and successful plotting.