Problem 32 - Algebra and Trigonometry

Question

\(23-34\) 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 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 at (4, 5) and (6, 5), and foci at \((5-\sqrt{2}, 5)\) and \((5+\sqrt{2}, 5)\).

Step 1: Rearrange and Identify

First, let's rearrange the given equation: \[ x^2 - y^2 = 10(x - y) + 1 \]Rewriting it, we have:\[ x^2 - 10x - y^2 + 10y = 1 \]We can see that this equation includes both \(x^2\) and \(y^2\), which suggests it could represent a conic section.

Step 2: Completing the Square

To simplify, complete the square for both \(x\) and \(y\):For \(x\): The equation involves \(x^2 - 10x\). To complete the square, take \(-\frac{10}{2}\) and square it: \((\frac{-10}{2})^2 = 25\). Add and subtract 25:\[ x^2 - 10x + 25 - 25 \]Becomes:\[ (x-5)^2 - 25 \]For \(y\): The equation involves \(-y^2 + 10y\). Complete the square here similarly: \((\frac{10}{2})^2 = 25\). Add and subtract 25:\[ -y^2 + 10y - 25 + 25 \]Becomes:\[ -(y-5)^2 + 25 \]

Step 3: Substitute Back into Equation

Substitute the completed square forms back into the main equation:\[ (x-5)^2 - 25 - ((y-5)^2 - 25) = 1 \]Simplify this:\[ (x-5)^2 - (y-5)^2 = 1 \]

Step 4: Identify the Conic Section

The simplified equation \((x-5)^2 - (y-5)^2 = 1\) fits the standard form of a hyperbola:\[ (x-h)^2 - (y-k)^2 = 1 \]This confirms the conic is a hyperbola, with the center at \((h, k) = (5, 5)\).

Step 5: Find Properties of the Hyperbola

Next, identify key properties of the hyperbola:- **Center**: \((5, 5)\)- **Vertices**: Since \(\pm 1\) is not scaled by another factor, vertices can be found by moving 1 unit from the center horizontally: - Vertices at \((4, 5)\) and \((6, 5)\)- **Foci**: Using \(c^2 = a^2 + b^2\) where \(a = 1\), \(b = 1\), \(c = \sqrt{2}\): - Foci at \((5-\sqrt{2}, 5)\) and \((5+\sqrt{2}, 5)\)- **Asymptotes**: They are given by the equation \( y - 5 = \pm (x - 5)\)

Key Concepts

Completing the Square

Completing the square is a technique used to simplify quadratic expressions and make them more manageable. This method transforms a quadratic expression into a perfect square trinomial, which reveals important features like vertices or centers, making it easier to identify the type of conic section represented by the equation.

To complete the square for a binomial like \(x^2 - 10x\), you take half of the coefficient of \(x\), square it, and then add and subtract this square from the expression. In this example, \((-\frac{10}{2})^2\) equals 25. Thus, \(x^2 - 10x\) becomes \((x-5)^2 - 25\).

For the \(y\) term, we have \(-y^2 + 10y\). Applying the same method gives \((\frac{10}{2})^2 = 25\). This means the expression is rewritten as \(-(y-5)^2 + 25\).

Completing the square allows us to convert a given equation into a recognizable form, revealing its geometric identity, and making further analysis of conic sections more straightforward.

Hyperbola

A hyperbola is a specific type of conic section characterized by its two separate curves, called branches. Hyperbolas form when a double cone is sliced perpendicular to its vertical axis. The standard equation for a hyperbola centered at \((h, k)\) is \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\) for horizontally oriented hyperbolas.

In our exercise, once we completed the square, the equation \((x-5)^2 - (y-5)^2 = 1\) appeared. It is in the standard form of a hyperbola, confirming that the conic section is indeed a hyperbola with the center at \((5, 5)\).

Hyperbolas have several distinct features:

• **Two branches** that open in opposite directions.
• They have asymptotes, which are lines the branches approach but never touch.
• A center, which is the midpoint between the vertices.
• Hyperbolas also differ from ellipses with a subtraction sign in their equation.

Vertices of Conic Sections

Vertices are key points on conic sections. For a hyperbola, the vertices are the points closest to the center on each branch. To find them, we look at the \(a^2\) value from the standard form of the hyperbola's equation.

In the equation \((x-5)^2 - (y-5)^2 = 1\), we identify \(a^2 = 1\) and \(a = 1\). This value of \(a\) tells us the distance from the center to each vertex in the x-direction because the hyperbola is horizontal.

Thus, starting from the center \((5, 5)\), we move 1 unit left and 1 unit right along the x-axis, resulting in vertices at \((4, 5)\) and \((6, 5)\). These points form the closest points of each branch of the hyperbola to its center.

Foci of Hyperbola

The foci of a hyperbola are crucial points located inside each branch. They are essential for defining a hyperbola, as the difference in distances from any point on the hyperbola to the foci is constant.

To find the foci for a hyperbola, we use the equation \(c^2 = a^2 + b^2\). Here, \(a\) and \(b\) are derived from the standard hyperbola equation. In our case, both \(a\) and \(b\) are 1. Therefore, \(c^2 = 1 + 1\) implying \(c = \sqrt{2}\).

Starting from the center at \((5, 5)\), we move \(\sqrt{2} \approx 1.41\) units left and right along the x-axis, giving us the foci at \((5-\sqrt{2}, 5)\) and \((5+\sqrt{2}, 5)\). These are pivotal points around which the branches of the hyperbola extend.