Problem 14 - Precalculus: Functions and Graphs

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

Vertices: \((-\frac{1}{2}, 1)\) and \((\frac{1}{2}, 1)\); Foci: \((0, 1 \pm \frac{\sqrt{3}}{2})\).

Step 1: Write the equation in standard form

First, start by rearranging the given equation to the form of an ellipse. The original equation is \(4x^2 + y^2 = 2y\). Completing the square for the \(y\) term, we obtain \(y^2 - 2y = (y-1)^2 - 1\), making the equation \(4x^2 + (y-1)^2 = 1\). Finally, divide through by 1 to achieve the standard form: \(\frac{x^2}{\frac{1}{4}} + \frac{(y-1)^2}{1} = 1\).

Step 2: Identify the key parameters

Compare the equation \(\frac{x^2}{\frac{1}{4}} + \frac{(y-1)^2}{1} = 1\) with the general form \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\) to identify parameters. Here, \(h = 0\), \(k = 1\), \(a^2 = \frac{1}{4}\) and \(b^2 = 1\), thus \(a = \frac{1}{2}\) and \(b = 1\).

Step 3: Determine the vertices

For an ellipse centered at point \((h, k)\), the vertices along the \(x\)-axis direction are \((h \pm a, k)\) and along the \(y\)-axis direction are \((h, k \pm b)\). Calculating these gives the vertices at \((0 \pm \frac{1}{2}, 1)\) which are \((-\frac{1}{2}, 1)\) and \((\frac{1}{2}, 1)\).

Step 4: Determine the foci

For an ellipse, the distance to the foci along the major axis is calculated by \(c = \sqrt{b^2 - a^2}\). Given \(b^2 = 1\) and \(a^2 = \frac{1}{4}\), we find \(c = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}\). The foci are at \((h, k \pm c)\) which are \((0, 1 \pm \frac{\sqrt{3}}{2})\), resulting in \((0, 1 + \frac{\sqrt{3}}{2})\) and \((0, 1 - \frac{\sqrt{3}}{2})\).

Step 5: Sketch the ellipse

Draw a coordinate system and plot the vertices \((-\frac{1}{2}, 1)\), \((\frac{1}{2}, 1)\), and the foci \((0, 1 + \frac{\sqrt{3}}{2})\) and \((0, 1 - \frac{\sqrt{3}}{2})\). Draw an ellipse that passes through the vertices, placing the foci within the ellipse as points along the vertical axis.

Key Concepts

vertices of an ellipse

The vertices of an ellipse are special points located at the intersection of the ellipse with its major axis. Imagine slicing the ellipse along its longest stretch, the vertices are where the ellipse meets the slice.
For a horizontal ellipse, these points lie along the x-axis, and for a vertical ellipse, as in our given ellipse, they lie along the y-axis.

To determine these points, we use the center of the ellipse \((h, k)\), and the distances \((\pm a, 0)\) and \((0, \pm b)\), where \('a'\) and \('b'\) measure the half-lengths of the major and minor axes.

• For the ellipse \[\frac{x^2}{\frac{1}{4}} + \frac{(y-1)^2}{1} = 1\], the center is at \( (0, 1) \).
• The vertices along the y-axis are calculated using \( b = 1 \), giving the vertices at \( (0, 1 + 1) = (0, 2) \) and \( (0, 1 - 1) = (0, 0) \).

Therefore, you plot the vertices at these calculated points to visualize the structure of the ellipse.

foci of an ellipse

Foci are two significant points inside each ellipse. They are used to define and understand the properties of an ellipse. The sum of the distances from any point on the ellipse to each focus is constant.
This unique property helps to form the shape of an ellipse. In our case, the foci lie on the same axis as the vertices but are not on the ellipse itself.

To find the foci positions, we use the equation \(c = \sqrt{b^2 - a^2}\), where \(c\) is the distance from the center to each focus.

• Given \( b^2 = 1, a^2 = \frac{1}{4} \), we find \( c = \sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2} \).
• The foci are then located at \( (h, k + c) \) and \( (h, k - c) \), which simplifies to \( (0, 1 + \frac{\sqrt{3}}{2}) \) and \( (0, 1 - \frac{\sqrt{3}}{2}) \).

These points should be marked accurately within the ellipse, following the vertical line through the center.

standard form of an ellipse

The standard form of an ellipse is crucial for identifying its dimensions and for plotting it correctly. An ellipse's equation is written as: \[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \].
This formula describes an ellipse centered at \(h, k\) with horizontal and vertical radii of \(a\) and \(b\).

By converting \(4x^2 + y^2 = 2y \) to standard form, we find:

• Centre coordinates as \( h = 0 \) and \( k = 1 \).
• Values \( a^2 = \frac{1}{4} \) (major axis radius squared) and \( b^2 = 1 \) (minor axis radius squared).

The use of completing the square helps transition it into this form, revealing the ellipse's orientations and dimensions when sketched.

completing the square

Completing the square is a vital algebraic technique used to convert quadratic equations into a form that is easier to work with, especially for ellipses.
It involves making a quadratic expression into a perfect square trinomial, which then leads to the standard form of an ellipsis equation.

In the original equation, \(4x^2 + y^2 = 2y\), the step-by-step process involves repositioning and adjusting terms:

• First, rearrange the equation to separate the y-terms: \( y^2 - 2y \).
• Then, use the identity \( y^2 - 2y = (y-1)^2 - 1 \) to complete the square.
• Incorporate this into the original formula: \( 4x^2 + (y-1)^2 = 1 \), leading to the standard form.
  

Through this process, we uncover a more descriptive equation of the ellipse that discloses its properties vividly.