Problem 49 - Precalculus

Question

Use a graphing utility to graph the ellipse. Find the center, foci, and vertices. (Recall that it may be necessary to solve the equation for \(y\) and obtain two equations.) $$5 x^{2}+3 y^{2}=15$$

Step-by-Step Solution

Answer

The center of the ellipse is at (0,0), the vertices are at (0, -sqrt(5)) and (0, sqrt(5)), and the foci are at (0, -sqrt(2)) and (0, sqrt(2)).

Step 1: Rewrite the Equation in Standard Form

We will start by rewriting the given equation \(5x^2 + 3y^2 = 15\) in the standard form. We can do this by dividing all terms by 15, The resulting equation will be: \(\frac{x^2}{3} + \frac{y^2}{5} = 1\)

Step 2: Identify the Center

By comparing the equation in standard form with the general standard form we can see that h=0 and k=0. Therefore, the center of the ellipse is at the origin, (0, 0).

Step 3: Calculate the Lengths of the Axes

Given that a and b represent the square root of the denominator of the standard form equation, we can find that a = sqrt(3) and b = sqrt(5). The semi-major axis is the larger of the two, which is b = sqrt(5), and the semi-minor axis is a = sqrt(3).

Step 4: Identify the Vertices

The vertices are located at (0, ±b), therefore they are (0, ±sqrt(5)).

Step 5: Calculate Foci

The foci are located at (0, ±c) where c = sqrt(b^2 - a^2). By substituting we get c = sqrt(5 - 3) = sqrt(2). Therefore, the foci are at (0, ±sqrt(2)).

Key Concepts

Standard Form of Ellipse Equation

Understanding the standard form of an ellipse equation is imperative for graphing and analyzing its properties. An ellipse's equation in standard form is expressed as \[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\]where \((h, k)\) represents the center of the ellipse, and \(a\) and \(b\) are the lengths of the semi-major and semi-minor axes, respectively. The denominators of these fractions, \(a^2\) and \(b^2\), determine the shape and size of the ellipse.

For horizontal ellipses, \(a^2\) is under the \(x\)-term and \(b^2\) under the \(y\)-term, while for vertical ellipses, this is reversed. In our example, after dividing each term by 15, the equation \(5x^2 + 3y^2 = 15\) becomes \[\frac{x^2}{3} + \frac{y^2}{5} = 1\], which is now in the standard form. The numerators are squared terms of \((x-h)\) and \((y-k)\), indicating that our ellipse is centered at the origin \((0, 0)\).

Vertices of an Ellipse

Vertices are critical points that define the boundary of an ellipse along its major axis. For an ellipse centered at the origin, the vertices can be found by examining the \(a\) and \(b\) values in the standard form equation. The vertices are either horizontally or vertically aligned depending on whether the ellipse is wider or taller.

The general position of the vertices for a horizontal ellipse is given by \((h±a, k)\), and for a vertical ellipse by \((h, k±b)\). In our exercise, since the center is at the origin \((0,0)\) and \(b = \sqrt{5}\), which is greater than \(a\), we determine that the ellipse is vertical and the vertices are therefore at \[(0, ±\sqrt{5}).\] These points represent the farthest north and south points on the ellipse from its center.

Foci of an Ellipse

The foci (plural of focus) of an ellipse are two distinct points located along the major axis, which have the unique property that the sum of the distances from the foci to any point on the ellipse is constant. For an ellipse in standard form centered at the origin, the foci are found using the formula \[c = \sqrt{a^2 - b^2}\] if \(a\) is greater than \(b\), and \[c = \sqrt{b^2 - a^2}\] if \(b\) is greater than \(a\). The coordinates of the foci are then \((h+c, k)\) and \((h-c, k)\) for a horizontal ellipse, or \((h, k+c)\) and \((h, k-c)\) for a vertical ellipse.

In the given exercise, since \(b > a\), we compute \(c = \sqrt{5 - 3} = \sqrt{2}\). As a result, the foci of this vertical ellipse are located at \[(0, ±\sqrt{2}).\]

Conic Sections

Ellipses are one of the four basic types of conic sections, which include circles, parabolas, and hyperbolas. These shapes are formed by the intersection of a plane with a double-napped cone. The angle of the intersecting plane relative to the cone's surface dictates the type of conic section produced. An ellipse is formed when the plane cuts through the cone at an angle, such that it intersects both naps of the cone but does so in a manner that is not perpendicular to the cone's axis.

Ellipses have unique mathematical properties and applications, ranging from planetary orbits to the design of whispering galleries. In algebra and geometry, learning to graph and analyze ellipses helps students gain a deeper understanding of conic sections and the broader world of mathematics and science.