Problem 14 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-14: Find the vertices and foci of the ellipse. Sketch its graph, showing the foci. $$ 4 x^{2}+y^{2}=2 y $$

Step-by-Step Solution

Answer

The vertices are at \((0, 2)\) and \((0, 0)\). The foci are at \((0, 1+\frac{\sqrt{3}}{2})\) and \((0, 1-\frac{\sqrt{3}}{2})\).

Step 1: Rearrange the Equation

Given the equation \(4x^2 + y^2 = 2y\), let's first rearrange it by moving the \(2y\) term to the left side. This gives us \(4x^2 + y^2 - 2y = 0\).

Step 2: Complete the Square for y

Complete the square for the \(y\) term in the equation \(4x^2 + y^2 - 2y = 0\). Rewrite \(y^2 - 2y\) as \((y-1)^2 - 1\). This transforms the equation to \(4x^2 + (y-1)^2 = 1\).

Step 3: Divide by Constant to Get Standard Form

Divide the entire equation \(4x^2 + (y-1)^2 = 1\) by 1 (which is the constant on the right side) to express it in the standard ellipse form: \(\frac{x^2}{rac{1}{4}} + \frac{(y-1)^2}{1} = 1\). This can be rewritten as \(\frac{x^2}{\left(\frac{1}{2}\right)^2} + \frac{(y-1)^2}{1} = 1\).

Step 4: Identify Vertices

The standard form \(\frac{x^2}{\left(\frac{1}{2}\right)^2} + \frac{(y-1)^2}{1} = 1\) identifies an ellipse centered at \((0, 1)\) with semi-major axis along the y-axis of length 1, and semi-minor axis along the x-axis of length \(\frac{1}{2}\). The vertices are at \((0, 1 + 1)\), and \((0, 1 - 1)\), which are \((0, 2)\) and \((0, 0)\).

Step 5: Establish Foci

For the given ellipse, the distance formula for the foci is \(c = \sqrt{a^2 - b^2}\), where \(a = 1\) and \(b = \frac{1}{2}\). Thus, \(c = \sqrt{1 - \left(\frac{1}{2}\right)^2} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}\). The foci lie along the y-axis at \((0, 1 + \frac{\sqrt{3}}{2})\) and \((0, 1 - \frac{\sqrt{3}}{2})\).

Step 6: Draw the Graph

Sketch the ellipse centered at \((0,1)\) with vertices at \((0, 2)\) and \((0, 0)\), and foci at \((0, 1 + \frac{\sqrt{3}}{2})\) and \((0, 1 - \frac{\sqrt{3}}{2})\). The semi-major axis runs vertically, and the semi-minor axis runs horizontally.

Key Concepts

Vertices of an Ellipse

The vertices of an ellipse are key reference points that help us understand the shape and orientation of the ellipse. To find the vertices, we start from the standard form equation of an ellipse, which looks like this: \[ \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1 \] Here,

• \((h, k)\) is the center of the ellipse,
• \(a\) is the semi-major axis if \(a > b\),
• The length of the semi-major axis is \(2a\),
• \(b\) is the semi-minor axis if \(a > b\).

In the given equation \[ \frac{x^2}{\left(\frac{1}{2}\right)^2} + \frac{(y-1)^2}{1} = 1 \], the center is at \((0, 1)\). As this is a vertically oriented ellipse, \(b = 1\) is the semi-major axis, and \(a = \frac{1}{2}\) is the semi-minor axis. Thus, the vertices are at \((0, 1 + 1)\) and \((0, 1 - 1)\), simplifying to \((0, 2)\) and \((0, 0)\).These vertices indicate the topmost and bottommost points on the ellipse aligned along the y-axis, providing a clear understanding of the ellipse's vertical stretch.

Foci of an Ellipse

The foci of an ellipse are unique points such that the sum of the distances from any point on the ellipse to each focus is constant. This property is fundamental in defining the geometric nature of an ellipse.Given the standard form equation mentioned earlier, the foci are located along the major axis at a distance \(c\) from the center. This distance \(c\) is calculated as:\[ c = \sqrt{a^2 - b^2} \]For our specific ellipse:

• \(a = \frac{1}{2}\): semi-minor axis
• \(b = 1\): semi-major axis (since the ellipse is vertical)

Thus, \[ c = \sqrt{1 - \left(\frac{1}{2}\right)^2} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]The focus points, aligned along the y-axis, are at:

• \((0, 1 + \frac{\sqrt{3}}{2})\)
• \((0, 1 - \frac{\sqrt{3}}{2})\)

These locations are inside the bounds set by the vertices, making them essential for understanding the "pull" that shapes the oval nature of the ellipse.

Completing the Square

Completing the square is a transformation technique used to convert a quadratic expression into a perfect square trinomial. This process allows us to reformat expressions so they fit the standard equation of shapes such as ellipses and circles.In our given problem, the equation started as:\[ 4x^2 + y^2 = 2y \]To transform it, we focus on the quadratic terms in \(y\)}. Rearranging, we have:\[ 4x^2 + y^2 - 2y = 0 \]To complete the square for the \(y\) terms:

• elliminate the term \(2y\)
• by adding and subtracting \(1\) inside the equation

This gives us:\[(y - 1)^2 - 1\]Thus, the equation is transformed into:\[ 4x^2 + (y - 1)^2 = 1 \]By rearranging and dividing through by the appropriate constants, we achieve the standard form of the ellipse, \[ \frac{x^2}{\left(\frac{1}{2}\right)^2} + \frac{(y-1)^2}{1} = 1 \]. Completing the square helps in identifying important properties such as center, axes, vertices, and foci, making it a crucial step in solving ellipse problems effectively.