Problem 13 - Calculus Early Transcendentals

Question

Find the vertices and foci of the ellipse and sketch its graph. $$4 x^{2}+y^{2}=16$$

Step-by-Step Solution

Answer

Vertices: (0, 4), (0, -4), (2, 0), (-2, 0); Foci: (0, 2√3), (0, -2√3).

Step 1: Write the Equation in Standard Form

The given equation is \(4x^2 + y^2 = 16\). To write it in standard form, divide every term by 16 to get \(\frac{x^2}{4} + \frac{y^2}{16} = 1\). It is now in the standard form of an ellipse centered at the origin: \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), where \(a^2 = 4\) and \(b^2 = 16\).

Step 2: Identify the Values of a and b

From the equation \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), we identify that \(a^2 = 4\) and \(b^2 = 16\). Therefore, \(a = 2\) and \(b = 4\). This tells us the lengths of the semi-major and semi-minor axes.

Step 3: Determine the Orientation of the Ellipse

Since \(b^2 > a^2\), the ellipse is vertically oriented. Hence, the longer axis (major axis) is aligned along the y-axis, and the shorter axis (minor axis) is along the x-axis.

Step 4: Find the Vertices

For a vertically oriented ellipse centered at the origin, the vertices are at \((0, \pm b)\) and \((\pm a, 0)\). So, the vertices are \((0, 4)\), \((0, -4)\), \((2, 0)\), and \((-2, 0)\).

Step 5: Calculate the Foci

The formula to find the foci of an ellipse is \((0, \pm c)\) for vertically oriented ellipses, where \(c = \sqrt{b^2 - a^2}\). Here, \(c = \sqrt{16 - 4} = \sqrt{12} = 2\sqrt{3}\). Thus, the foci are \((0, 2\sqrt{3})\) and \((0, -2\sqrt{3})\).

Step 6: Sketch the Ellipse

Sketch the ellipse by plotting the vertices and foci determined in previous steps. The ellipse's major axis is vertical along the y-axis, passing through the points \((0, 4)\) and \((0, -4)\), and the minor axis is horizontal along the x-axis passing through \((2, 0)\) and \((-2, 0)\). The foci \((0, 2\sqrt{3})\) and \((0, -2\sqrt{3})\) lie inside the ellipse along the major axis.

Key Concepts

Standard form of an ellipse

The standard form of an ellipse is a vital concept that offers a structured way to evaluate and explore the properties of an ellipse. By expressing the ellipse equation \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]we identify the ellipse's dimensions and orientation easily. In this form, the ellipse is centered at the origin, where:

• \(a^2\) and \(b^2\) represent the squares of the lengths of the semi-major and semi-minor axes.
• Notice that \(a\) is always associated with the x-term and \(b\) with the y-term when the ellipse is horizontally oriented, whereas they swap roles in a vertically oriented ellipse.

In our example, the initial equation \(4x^2 + y^2 = 16\) was transformed by dividing each term by 16, leading to the standard form \(\frac{x^2}{4} + \frac{y^2}{16} = 1\). This standard form allows us to quickly retrieve information about the ellipse's properties.

Vertices of an ellipse

Vertices are prominent points located on the major axis of an ellipse, serving as the extremities. Determining the vertices helps define the shape in which the ellipse is drawn. For ellipses centered at the origin, the vertices lie at the points:

• \( (0, \pm b) \) when the ellipse is vertically oriented.
• \( (\pm a, 0) \) reflects its position along smaller dimensions if it's horizontally oriented.

In the problem provided, the resulting vertices were calculated as:

• \( (0, 4) \) and \( (0, -4) \), aligning with the longer, vertical axis.
• \( (2, 0) \) and \( (-2, 0) \), belonging to the shorter, horizontal axis.

Each vertex shows where the ellipse extends the furthest, framing the boundary for potential sketches.

Foci of an ellipse

The foci of an ellipse are unique interior points that carry distinctive geometrical importance, particularly in terms of how the ellipse is defined mathematically. The distance from each point on the ellipse to the foci is constant when added together. The location of the foci is given as:

• \( (0, \pm c) \) for vertically oriented ellipses.
• The formula for \(c\) is \( c = \sqrt{b^2 - a^2} \), revealing how foci depend on the axes' lengths.

For the solution, where \( b^2 = 16 \) and \( a^2 = 4 \):

• Calculate \( c = \sqrt{16 - 4} = \sqrt{12} = 2\sqrt{3} \).
• The foci are positioned at \( (0, 2\sqrt{3}) \) and \( (0, -2\sqrt{3}) \).

Understanding the foci's locations enriches insight into an ellipse's dimensional attributes and helps complete the sketch.

Equation of an ellipse

The equation of an ellipse is not just a mathematical representation but a tool that unearths the essential components of its structure, motivating comprehensive exploration. An ellipse equation in the general form \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]reveals the balance of the axes:

• This generalized format ensures the center is at the origin and makes it easy to determine the orientation and size by analyzing \(a\) and \(b\).
• For ellipses aligned with the coordinate axes, the specific contribution of each variable is visible.

In the initial step of transforming \(4x^2 + y^2 = 16\), into this form, by dividing through by 16, we find the meaningful expression, \[ \frac{x^2}{4} + \frac{y^2}{16} = 1, \]indicating a vertically oriented ellipse with clearly defined axes lengths and orientation. This process presents how manipulation of algebraic expressions guides us toward the visualization and interpretation of complex, geometric shapes.