Problem 26 - Precalculus: Functions and Graphs

Question

Exer \(19-36:\) Find an equation for the ellipse that has its center at the origin and satisfies the given conditions. Vertices \(M \pm 13,0), \quad\) passing through \((5,6)\)

Step-by-Step Solution

Answer

The equation is \( \frac{x^2}{169} + \frac{y^2}{42.25} = 1 \).

Step 1: Understand the Form of the Ellipse Equation

An ellipse centered at the origin (0,0) typically has the equation \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) if the major axis is along the x-axis. Here, \(a\) is the semi-major axis length, and \(b\) is the semi-minor axis length.

Step 2: Determine the Semi-Major Axis Length (a)

Given the vertices at \((\pm 13,0)\), the length of the major axis is 26. The semi-major axis \(a\) is half the length of the major axis, so \(a = 13\).

Step 3: Write the Partial Equation of the Ellipse

Substitute \(a = 13\) into the ellipse equation. The equation now becomes \(\frac{x^2}{13^2} + \frac{y^2}{b^2} = 1\).

Step 4: Use Given Point to Find b²

Since the point (5,6) lies on the ellipse, substitute \(x = 5\) and \(y = 6\) into the partial equation: \(\frac{5^2}{13^2} + \frac{6^2}{b^2} = 1\). This simplifies to \(\frac{25}{169} + \frac{36}{b^2} = 1\).

Step 5: Solve for b²

Rearrange the equation from Step 4 to solve for \(b^2\): \(\frac{36}{b^2} = 1 - \frac{25}{169}\). This simplifies to \(\frac{36}{b^2} = \frac{144}{169}\). Therefore, \(b^2 = \frac{36 \times 169}{144} = 42.25\).

Step 6: Write the Final Equation of the Ellipse

Now that we have the values for \(a^2 = 169\) and \(b^2 = 42.25\), the equation of the ellipse is \(\frac{x^2}{169} + \frac{y^2}{42.25} = 1\).

Key Concepts

Semi-Major Axis

In the equation of an ellipse, the semi-major axis is a crucial element. It defines the longest radius from the center of the ellipse to its edge. This distance is what distinguishes a stretched-out ellipse from a perfect circle. A longer semi-major axis means a more elongated shape.

For our ellipse, we know from the vertices given as \(( ext{M} \pm 13,0)\) that the total length of the major axis is 26. Since the semi-major axis is half of this total length, we calculate it as follows:

• Length of major axis = 26
• Semi-major axis, \(a = \frac{26}{2} = 13\)

This value can then be plugged into the standard form of the ellipse equation to reflect the specific dimensions of our ellipse.

Ellipse Center at Origin

When an ellipse is described with its center at the origin, it simplifies the equation significantly. The origin, denoted as \((0,0)\), acts as a reference point where the ellipse is symmetrical.

In simpler terms:

• The center at the origin implies the ellipse is equally distributed around this point, with no shift
• This symmetry allows us to use the simpler form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\)

The equation is much easier to tackle because there's no need for additional variables to account for any shifts in the x or y direction, which would be present if the center was not at the origin. This standardization makes calculations more straightforward, as seen in simplifying the form when substituting values.

Equation of Conic Sections

Ellipses are part of a larger family called conic sections. These are curves obtained by the intersection of a plane with a double-napped cone, which also includes circles, parabolas, and hyperbolas. For an ellipse located at the origin, the equation is simplified into a canonical form.

• Standard form: \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\)
• Here, \(a\) and \(b\) represent the semi-major and semi-minor axis respectively

To finalize our specific ellipse equation from the provided problem, substitute the determined values of \(a^2 = 169\) and \(b^2 = 42.25\) into this formula. Thus, we derive \(\frac{x^2}{169} + \frac{y^2}{42.25} = 1\), efficiently describing our conic section in standard form. This systematic approach helps demystify the process and demonstrates the application of conic section principles.