Problem 25 - Precalculus with Limits

Question

In Exercises 19-28, find the standard form of the equation of the ellipse with the given characteristics. Center: \((0, 4); \quad a=2c; \quad\) vertex: \((-4, 4), (4, 4)\)

Step-by-Step Solution

Answer

The standard form of the ellipse with the given characteristics is \(x^2 / 16 + (y-4)^2 / 12 = 1\).

Step 1: Understand Ellipse Characteristics

The ellipse is defined by the equation \[(x-h)^2 / a^2 + (y-k)^2 / b^2 = 1\] where (h, k) is the center of the ellipse, a is the distance from the center to a vertex, and b is the distance from the center to a co-vertex. In this problem, it is also given that \(a = 2c\), where c is the distance from the center to a focus.

Step 2: Find Values of a, b, and c

The center of the ellipse is given as (0, 4). The vertex points are given as (-4, 4) and (4,4), so the value of a (distance from the center to a vertex) can be found as \( a = 4 - 0 = 4\). Being that \(a = 2c\), then \(c = a/2 = 2\). To find value b, recall that in an ellipse \(a^2 = b^2 + c^2\). Substituting known values, we can solve the equation \( b^2 = a^2 - c^2 = 4^2 - 2^2 = 12\), and since b is a distance, we take its positive root, \( b = sqrt{12} = 2sqrt{3}\).

Step 3: Plug Values into Ellipse Formula

Finally, with these known values, plug into the standard form of the equation for the ellipse to find \[(x-h)^2 / a^2 + (y-k)^2 / b^2 = 1\] becomes \[(x-0)^2 / 16 + (y-4)^2 / 12 = 1\], which simplifies to \[x^2 / 16 + (y-4)^2 / 12 = 1\].

Key Concepts

Center of Ellipse

The center of an ellipse is a crucial element in understanding its geometry. It acts as the reference point from which distances to vertices and foci are measured. Given in the exercise, the center is

• (0, 4). This means the ellipse is centered horizontally along the x-axis, and vertically four units above the origin.

Knowing the center helps in writing the standard form equation and visualizing the ellipse's position on a coordinate plane, making calculations easier and more intuitive.

Distance from Center to Vertex

The distance from the center of the ellipse to a vertex, denoted as \(a\), is one of the key measures that defines the ellipse's size and shape. Vertices represent the points where the ellipse is widest or narrowest, depending on its orientation.For the given exercise:

• Vertices are (-4, 4) and (4, 4), showing the ellipse is horizontal.
• The distance \(a\) is calculated as the horizontal distance from (0, 4) to one of these vertices which equals 4 units.

This distance helps determine other essential parameters like the distance to the foci, allowing further equations to be solved.

Standard Form of Ellipse

The standard form of the ellipse equation is vital for expressing its characteristics mathematically. The equation \[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\]uses \(h\) and \(k\) as the center coordinates, \(a\) as the semimajor axis, and \(b\) as the semiminor axis.For this problem:

• \((h, k) = (0, 4), \)
• \(a = 4, \quad c = 2 \quad (a = 2c),\)
• \(b = \sqrt{12} = 2\sqrt{3}.\)

Plug these into the formula to get the equation: \[\frac{x^2}{16} + \frac{(y-4)^2}{12} = 1\]This standard form is essential for calculation and graphing, giving insight into the ellipse's orientation, scale, and relation to its center.