Problem 18 - Algebra 2

Question

Write each equation in standard form. State whether the graph of the equation is a parabola, circle, ellipse, or hyperbola. Then graph the equation. $$ x^{2}-y^{2}+8 x=16 $$

Step-by-Step Solution

Answer

It's a hyperbola centered at (-4, 0) with vertices at (-8, 0) and (0, 0).

Step 1: Move all terms to one side

Begin by rearranging the given equation to bring all terms to one side. The original equation is \( x^2 - y^2 + 8x = 16 \). Subtract 16 from both sides to get the equation in standard mathematical form: \( x^2 - y^2 + 8x - 16 = 0 \).

Step 2: Complete the square for the x-term

To transform the equation into a recognizable conic section, complete the square on the \( x \)-part. Starting with \( x^2 + 8x \), we take half of the coefficient of \( x \), which is 4, square it to get 16, and add & subtract it in the equation:\( (x^2 + 8x + 16 - 16) - y^2 - 16 = 0 \). This becomes \((x + 4)^2 - y^2 = 16 \).

Step 3: Identify the conic section

With the equation in the form \((x + 4)^2 - y^2 = 16\), observe that it is a standard form of a hyperbola \( (x - h)^2 - (y - k)^2 = a^2 \), where \((h, k)\) is the center. Here, \(h = -4\), \(k = 0\), and \(a^2 = 16\).

Step 4: Sketch the graph of the hyperbola

Center the hyperbola at \((-4, 0)\). Since \((x + 4)^2 - y^2 = 16\) represents a horizontal hyperbola, the transverse axis is parallel to the x-axis. The vertices are at \((-4 - 4, 0) = (-8, 0)\) and \((-4 + 4, 0) = (0, 0)\). The equations of the asymptotes are \(y = \pm (x + 4)\). Draw these key features to sketch the hyperbola.

Key Concepts

Hyperbola

A hyperbola is a type of conic section that appears as two separate, mirror-image curves. It is distinct from other conic sections like circles, ellipses, and parabolas. The standard form for a hyperbola can take a horizontal or vertical orientation. For a horizontal hyperbola, the equation is \[\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\]While for a vertical hyperbola, it is \[\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\]where \(h, k\) represents the center of the hyperbola.

• The transverse axis has a length of \(2a\), and the vertices are along this axis.
• The conjugate axis is perpendicular to the transverse, with a length of \(2b\).
• Asymptotes for the hyperbola open out from the center, guiding how wide the branches open.

Understanding these components is critical to identify and graph hyperbolas accurately.

Standard Form

To convert the equation of a hyperbola to its standard form, you need to rearrange and simplify it until it looks like the standard form expressions outlined above. In our exercise, after completing the square on the x-term, the equation \[(x + 4)^2 - y^2 = 16\]resembles the standard form for a horizontal hyperbola. The presence of the minus sign between the squared terms indicates a hyperbola. The form gives us vital information about the hyperbola:

• The center \( (h, k) \) is \( (-4, 0) \).
• \(a^2 = 16\), suggesting \(a = 4\).
• Since there is no \(b^2\) term visible, if comparing, you'd think of the \(b^2\) being 1, though it's not necessarily needed without a \(y\) term adjusted.

By putting the equation into this form, you can identify and plot key features and accurately recognize the type of conic section represented by the equation.

Completing the Square

Completing the square is a technique used to simplify quadratic expressions or equations. It is especially helpful for rewriting conic sections' equations into standard form.To complete the square:

• For a quadratic expression, such as \(x^2 + bx\), identify \(b\), take half of \(b\), and square it.
• Add and subtract the squared value to/from the equation, creating a perfect square trinomial.

In our problem, the term \(x^2 + 8x\) undergoes this process:

• Half of 8 is 4, and squaring gives 16.
• Add and subtract 16, transforming the expression to \( (x + 4)^2 - 16 \).

This process turns the quadratic into a perfect square trinomial, making it easier to convert into the standard form for further analysis.

Graphing

Graphing a hyperbola involves plotting several key features based on its standard form. For the hyperbola \[(x + 4)^2 - y^2 = 16\], follow these steps to sketch the graph:

• Identify the center at \( (-4, 0) \).
• Since \(a^2 = 16\), find \(a = 4\). This helps locate the vertices at \((-4 - 4, 0)\) and \((-4 + 4, 0)\), or \((-8, 0)\) and \( (0, 0) \), being on the line y = 0.
• Draw the asymptotes through the center with equations \(y = \pm (x + 4)\), forming the diagonals of a rectangle centered at \( (-4, 0) \).

With these features identified, sketch smooth curves opening away from each other, giving you the characteristic hyperbola shape. The asymptotes serve as guidelines, showing how the hyperbola opens.