Problem 32 - Precalculus Mathematics for Calculus

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. $$x^{2}-y^{2}=10(x-y)+1$$

Step-by-Step Solution

Answer

The equation represents a hyperbola centered at (5, 5) with vertices (4, 5), (6, 5), and foci at (5 ± √2, 5).

Step 1: Rearrange the Equation

Start by rearranging the given equation so that all terms involving squares of \(x\) and \(y\) are together on one side: \[ x^2 - 10x - y^2 + 10y = 1 \].

Step 2: Complete the Square for x-terms

Consider the \(x\) terms: \(x^2 - 10x\). To complete the square, add and subtract \(\left(\frac{-10}{2}\right)^2 = 25\): \[ x^2 - 10x + 25 - 25 \]. This transforms into:\[(x - 5)^2 - 25\].

Step 3: Complete the Square for y-terms

Similarly, consider the \(y\) terms: \(-y^2 + 10y\). Factor out -1 to switch signs, as we are completing square on \(y\): \[- (y^2 - 10y)\]. Add and subtract \(\left(\frac{10}{2}\right)^2 = 25\): \[- (y^2 - 10y + 25 - 25) \]. This transforms into:\[- ((y - 5)^2 - 25) = - (y - 5)^2 + 25\].

Step 4: Substitute back into Equation

Substitute the complete square forms back into the equation:\[(x - 5)^2 - 25 - ( (y - 5)^2 - 25) = 1 \]. Simplify to:\[(x - 5)^2 - (y - 5)^2 = 1 \].

Step 5: Identify the Conic Section

The equation \[(x - 5)^2 - (y - 5)^2 = 1 \] is in the standard form of a hyperbola, which is \[(x - h)^2/a^2 - (y - k)^2/b^2 = 1 \]. This identifies the conic as a hyperbola.

Step 6: Determine Center, Vertices, and Foci

For the hyperbola \[(x - 5)^2 - (y - 5)^2 = 1 \]:- Center: (5, 5)- Transverse axis lies along the x-axis. The distance between center to vertices is \( a = 1 \).- Vertices are at: (5 + 1, 5) = (6, 5) and (5 - 1, 5) = (4, 5).- The foci distance from center is \( c = \sqrt{a^2 + b^2} = \sqrt{1^2 + 1^2} = \sqrt{2} \).- Foci are at: (5 + \sqrt{2}, 5) and (5 - \sqrt{2}, 5).The asymptotes are given by:\[ y - 5 = \pm 1(x - 5) \], or \( y = x \) and \( y = -x + 10 \).

Step 7: Sketch the Hyperbola

Draw the hyperbola centered at (5, 5), with vertices at (4, 5) and (6, 5). The foci should be at approximately (5 + \sqrt{2}, 5) and (5 - \sqrt{2}, 5). Sketch asymptotes which intersect at the center (5, 5), forming straight lines given by \( y = x \) and \( y = -x + 10 \).

Key Concepts

Completing the Square

To determine the nature of our conic section, we often use a technique called **completing the square**. This helps simplify quadratic expressions, making it easier to rewrite the general form of conic sections into their standard forms.

Completing the square involves manipulating a quadratic expression so that it becomes a perfect square trinomial. For a term like \( x^2 - bx \), we add and subtract \( \left(\frac{b}{2}\right)^2 \) to create a perfect square. Here's how it works in practice:

• Identify the \'x\' and \'y\' groups separately in your equation.
• For each group, determine the value necessary to complete the square.
• Add and subtract this value within each group to maintain equation balance.

Post completion, you can factor each side to express it as a squared binomial. In our example, by completing the squares for both \(x\) and \(y\), we transformed our equation into \[(x - 5)^2 - (y - 5)^2 = 1\]. This step is crucial to later steps, like identifying the type of conic section from the equation.

Hyperbola

A **hyperbola** is one of the four types of conic sections formed by the intersection of a plane and a double-napped cone. It resembles two mirrored, open-ended curves extending to infinity.

Mathematically, the equation of a hyperbola in its standard form involves a subtraction between two squared terms, and it looks like this: \[(x - h)^2/a^2 - (y - k)^2/b^2 = 1\]. Here, \((h, k)\) indicates the center of the hyperbola, and \(a\) and \(b\) represent distances that define the hyperbola's shape.

• A hyperbola has two separate branches.
• The minus sign indicates that the foci lie along the transverse axis.
• The equation's right side is always \(1\) when the hyperbola crosses the y-axis.

Understanding this equation is key to finding other characteristics like vertices, foci, and the asymptotic directions.

Vertices of a Hyperbola

In a hyperbola, the **vertices** are critical points on the graph where each branch of the hyperbola is closest to the center. They are aligned along the hyperbola's transverse axis, which is the main axis passing through the center, from one vertex to the other.

To find the vertices coordinates in an equation of the form \((x - h)^2/a^2 - (y - k)^2/b^2 = 1\), you simply move \(a\) units in either direction from the center \((h, k)\) along the transverse axis. For our equation:

• The center is at \((5, 5)\).
• Since \(a=1\), the vertices are located one unit left and right of the center.
• Thus, the vertices are at (6, 5) and (4, 5).

Plotting these points helps us to visualize the core structure of the hyperbola.

Foci of a Hyperbola

The **foci** are two important points located inside each branch of the hyperbola. They dictate the shape and the spacing of the branches around them.

To find the foci of a hyperbola given by the equation \((x - h)^2/a^2 - (y - k)^2/b^2 = 1\), use the following steps:

• Calculate the distance \(c\) from the center to each focus using the formula \( c = \sqrt{a^2 + b^2} \).
• For our problem, since both \(a\) and \(b\) equal 1, \(c = \sqrt{1^2 + 1^2} = \sqrt{2}\).
• The foci are positioned \(c\) units away from the center along the transverse axis.
• Thus, the foci of the hyperbola are \((5 + \sqrt{2}, 5)\) and \((5 - \sqrt{2}, 5)\).

The foci are often less visible on a graph but vital for understanding the hyperbola's geometry.

Asymptotes of a Hyperbola

The **asymptotes** of a hyperbola are straight lines that the hyperbola approaches as it extends towards infinity. They mark the directions in which the branches of the hyperbola head off.

For a hyperbola with its standard equation \((x - h)^2/a^2 - (y - k)^2/b^2 = 1\), the asymptotes are found using the equation:\( y - k = \pm \frac{b}{a} (x - h) \). This describes two intersecting lines that pass through the center of the hyperbola at \((h, k)\). For our hyperbola:

• The center is \( (5, 5) \).
• The slopes \( \pm \frac{1}{1} = \pm 1 \) indicate 45-degree angles.
• So, the asymptotes are \(y - 5 = \pm 1(x - 5)\), simplifying to \(y = x\) and \(y = -x + 10\).

These asymptotes give a visual frame within which the hyperbola's branches extend.