Problem 26 - Calculus Early Transcendentals: Pearson New International Edition

Question

Sketch the graph of the given equation. \(x^{2}-4 y^{2}-14 x-32 y-11=0\)

Step-by-Step Solution

Answer

Hyperbola, centered at (7, -4), with vertices at (7, -3) and (7, -5).

Step 1: Recognize the Form of the Equation

The given equation is \(x^2 - 4y^2 - 14x - 32y - 11 = 0\). Notice that this corresponds to the general form of a hyperbola, \(Ax^2 + By^2 + Cx + Dy + E = 0\), with \(A = 1\) and \(B = -4\). This indicates the equation is of a hyperbola type.

Step 2: Rearrange the Equation

Reorder the terms of the equation: \( x^2 - 14x - 4y^2 - 32y = 11 \). This makes it easier to complete the square for both \(x\) and \(y\).

Step 3: Complete the Square for x

Rewrite \(x^2 - 14x\) by completing the square: \(x^2 - 14x = (x-7)^2 - 49\). This turns it into a perfect square trinomial.

Step 4: Complete the Square for y

Rewrite \(-4y^2 - 32y\) by completing the square:\[-4(y^2 + 8y) = -4((y+4)^2 - 16) = -4(y + 4)^2 + 64\].

Step 5: Substitute Completed Squares into Original Equation

Place the completed squares back into the equation: \[(x-7)^2 - 49 - 4(y + 4)^2 + 64 = 11\]. Simplify to get \((x-7)^2 - 4(y+4)^2 = -4\).

Step 6: Adjust Equation for Hyperbola Standard Form

Divide all terms by \(-4\) to rearrange the equation to the standard form of a hyperbola: \[-\frac{(x-7)^2}{4} + (y+4)^2 = 1\]. This can also be arranged as: \[(y+4)^2 - \frac{(x-7)^2}{4} = 1\].

Step 7: Identify Hyperbola Characteristics

The standard form \((y-k)^2 / a^2 - (x-h)^2 / b^2 = 1\) indicates a vertical hyperbola centered at \((h, k)\) which here is \((7, -4)\). Here, \(a^2 = 1\) and \(b^2 = 4\), giving \(a = 1\) and \(b = 2\).

Step 8: Plot the Center, Vertices, and Asymptotes

Plot the center at \((7, -4)\). The vertices will be \(1\) unit up and down from the center along the y-axis at \((7, -3)\) and \((7, -5)\). The slopes of the asymptotes are \(\pm b/a = \pm 2\).

Key Concepts

Completing the Square

Completing the square is an essential technique when working with conic sections like hyperbolas. It's a method we use to transform a quadratic expression into a perfect square trinomial, making it much easier to manipulate and understand. For a quadratic in terms of

• \(x^2 - 14x\), we rearrange it as \((x-7)^2 - 49\)
• and for \(-4y^2 - 32y\), it becomes \(-4((y+4)^2 - 16)\).

This process involves identifying the coefficients, dividing them by two, squaring the result, and then adding and subtracting this square within the expression.
Doing this helps in simplifying the original quadratic into standard forms, where you can clearly see the transformation involved.

Standard Form of Hyperbola

The standard form of a hyperbola reveals the structure directly. It's represented as either

• \((x-h)^2/a^2 - (y-k)^2/b^2 = 1\)
• or \((y-k)^2/a^2 - (x-h)^2/b^2 = 1\).

This transformation is key when graphing because it showcases whether the hyperbola opens horizontally or vertically. For our given equation

• Completed the square rearranges it into: \( (y+4)^2 - \frac{(x-7)^2}{4} = 1\).

This translates into a vertical hyperbola, where the principal axis runs parallel to the y-axis. Recognizing this form allows you to find the center, vertices, and asymptotes more directly.

Asymptotes of Hyperbola

The asymptotes are crucial for sketching the graph of a hyperbola since they guide the shape of the branches. For a hyperbola

• \( (y-k)^2/a^2 - (x-h)^2/b^2 = 1\), the asymptotes will have equations \(y - k = \pm (b/a)(x - h)\).

These are straight lines passing through the center. For our particular hyperbola

• with \(a = 1\) and \(b = 2\), so the slope \(b/a = 2\),

giving us steeply sloped asymptotes. Asymptotes stretch indefinitely and provide a frame around which the hyperbola curves, never intersecting as it extends.
Understanding and plotting the asymptotes first can make drawing the rest of the hyperbola more intuitive.

Hyperbola Center and Vertices

The center of a hyperbola is like its anchor point, around which the entire graph is symmetrically structured. Determining this point comes directly from the standard form:

• \((h,k)\), represented easily in our converted equation as \((7, -4)\).

Once you have the center, finding the vertices is straightforward. They are central points on the hyperbola's principal axis, located at a distance equal to \(a\) from the center.
For a vertical hyperbola like ours

• the vertices will be \(1\) unit up and down from the center at \((7, -3)\) and \((7, -5)\).

These vertices are essential to carefully sketch the initial points of the hyperbola's branches, helping visually balance the hyperbolic shape on the graph.