Problem 24 - Algebra and Trigonometry

Question

Complete the square to determine whether the equation represents an ellipse, a parabola, a hyperbola, or a degenerate conic. If the graph is an ellipse, find the center, foci, vertices, and lengths of the major and minor axes. If it is a parabola, find the vertex, focus, and directrix. If it is a hyperbola, find the center, foci, vertices, and asymptotes. Then sketch the graph of the equation. If the equation has no graph, explain why. $$ 2 x^{2}+y^{2}=2 y+1 $$

Step-by-Step Solution

Answer

The equation represents an ellipse centered at (0, 1) with vertices at (0, 1 \pm \sqrt{2}), foci at (0, 0) and (0, 2), a major axis length of 2\sqrt{2}, and a minor axis length of 2.

Step 1: Rearrange the equation

First, we need to rearrange the equation to separate terms involving the same variable. Start by moving the linear term with y to the left side of the equation: \[ 2x^2 + y^2 - 2y = 1 \]

Step 2: Complete the square for the y-terms

To complete the square for the \( y \) terms, take the coefficient of \( y \), divide by 2, then square it, and finally add and subtract this value inside the equation. Here, the coefficient of \( y \) is \(-2\): \[ y^2 - 2y = (y-1)^2 - 1 \] Now substitute back into the equation: \[ 2x^2 + (y-1)^2 - 1 = 1 \]

Step 3: Simplify the equation

Add 1 to both sides to simplify: \[ 2x^2 + (y-1)^2 = 2 \]

Step 4: Normalize the equation

Divide every term by 2 to write the equation in a standard form: \[ \frac{x^2}{1} + \frac{(y-1)^2}{2} = 1 \] This equation is in the standard form for an ellipse.

Step 5: Identify elements of the ellipse

This is an ellipse centered at \((0, 1)\). The major axis is along the y-axis with semi-major axis length \( \sqrt{2} \) and the minor axis along the x-axis with semi-minor axis length \( 1 \).

Step 6: Determine foci and vertices of the ellipse

The foci are positioned along the major axis. Calculate the distance \( c \) using \( c = \sqrt{a^2 - b^2} = \sqrt{2 - 1} = 1 \). Thus, the foci are at \((0, 1 \pm 1) = (0, 0) \text{ and } (0, 2)\). The vertices are at \((0, 1 \pm \sqrt{2}) = (0, 1-\sqrt{2}) \text{ and } (0, 1+\sqrt{2})\).

Step 7: Summary and sketch

The given equation represents an ellipse centered at \((0, 1)\) with foci at \( (0, 0) \) and \( (0, 2) \), vertices at \( (0, 1 - \sqrt{2}), (0, 1 + \sqrt{2}) \), a major axis length of \(2\sqrt{2}\), and a minor axis length of 2.

Key Concepts

Ellipse

An ellipse is a type of conic section that you can think of as a stretched circle. It is shaped like an oval and has two main features: its center point and two axes. The longer axis is called the major axis, and the shorter one is the minor axis. An ellipse's shape will vary based on the lengths of these axes. If the major and minor axes are the same length, the ellipse looks like a perfect circle.
When you encounter an ellipse in an equation, it usually looks like \[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\] where \(h,k\) is the center of the ellipse. The values of \(a\) and \(b\) help determine the ellipse's shape by indicating the lengths of the semi-major and semi-minor axes. It's important to note that whichever of \(a^2\) or \(b^2\) is larger, decides which axis is the major one.

Completing the Square

Completing the square is a mathematical technique that helps to rewrite quadratic expressions into a more manageable form. This process is often essential when dealing with equations of conic sections like ellipses.
Here's how it generally works: given a quadratic expression like \(x^2 + bx\), you first take half of the coefficient of \(x\), square it, and then add and subtract this square within the expression. For example, if the expression is \(y^2 - 2y\):

• Take the coefficient of the \(y\) term, which is -2.
• Divide it by 2 to get -1.
• Square it to get 1.
• Rewrite as \(y^2 - 2y = (y - 1)^2 - 1\).

This method helps simplify the equation, making it easier to handle and identify characteristics of the conic section.

Equation of Ellipse

The equation of an ellipse is key in determining its properties and visualizing its graph. Once an equation is rewritten in its standard form, identifying these properties becomes straightforward. For example:\[\frac{x^2}{1} + \frac{(y-1)^2}{2} = 1\]in the format of \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\). This tells us:

• The center of the ellipse is at \( (0, 1) \).
• The semi-major axis and semi-minor axis are of lengths \( \sqrt{2} \) and \( 1 \) respectively.
• The major axis is vertical, as the \( (y-1)^2 \) term is divided by the larger number, indicating it stretches more along the y-axis.

Recognizing these characteristics is crucial for understanding the graph's shape and dimensions.

Vertices and Foci of Ellipse

The vertices and foci are vital features of an ellipse that help in visualizing and understanding its geometric properties.
Vertices are the points where the ellipse is wider or taller, lying on the major axis. For an equation like\[\frac{x^2}{1} + \frac{(y-1)^2}{2} = 1\],the vertices are located at \( (0, 1 \pm \sqrt{2}) \). The distance between the center and each vertex equals the semi-major axis length.
Foci, on the other hand, are located inside the ellipse along the major axis. They are points from which the sum of distances to any point on the ellipse is constant. The position of the foci can be calculated using the formula \( c = \sqrt{a^2 - b^2} \). For this ellipse, \( c = \sqrt{2 - 1} = 1 \), so the foci are at \( (0, 0) \) and \( (0, 2) \). These features play a crucial role in defining the specific shape and orientation of the ellipse.