Problem 82 - A Graphical Approach to Precalculus with Limits

Question

Write the equation in standard form for a hyperbola centered at ( \(h, k\) ). Identify the center and vertices. $$y^{2}+8 y-3 x^{2}+13=0$$

Step-by-Step Solution

Answer

Center: (0, -4); Vertices: (0, -3.73) and (0, -4.27).

Step 1: Group Terms

Let's start by rearranging the given equation, grouping the like terms together: y terms: \( y^2 + 8y \)x terms: \(-3x^2\).We want to write these grouped terms separately as: \[ y^2 + 8y - 3x^2 = -13 \]

Step 2: Complete the Square for y

To transform the \( y \) terms, we need to complete the square. Look at \( y^2 + 8y \). Half of 8 is 4, and 4 squared is 16.We add and subtract 16 to complete the square:\[ y^2 + 8y + 16 - 16 \]Rewrite as: \[ (y + 4)^2 - 16 \].

Step 3: Rewrite the Equation

Replace \( y^2 + 8y \) with \((y + 4)^2 - 16\) in the equation:\[ (y + 4)^2 - 16 - 3x^2 = -13 \] Simplify this to:\[ (y + 4)^2 - 3x^2 = 3 \].

Step 4: Convert to Standard Form of a Hyperbola

Divide the entire equation by 3 so the right side equals 1:\[ \frac{(y+4)^2}{3} - \frac{3x^2}{3} = \frac{3}{3} \]Simplify to:\[ \frac{(y+4)^2}{3} - x^2 = 1 \]This is the standard form of a hyperbola centered at \((0, -4)\) with the transverse axis parallel to the y-axis.

Step 5: Identify the Center

The equation is in the form \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \), which means the center is \((h, k)\).Here, \(h = 0\) and \(k = -4\), so the center is \((0, -4)\).

Step 6: Find the Vertices

For hyperbolas of form \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \), vertices are located \(a\) units above and below the center along the y-axis because the y-term is positive.Given \(\frac{(y+4)^2}{3}\), \(a^2 = 3\), so \(a = \sqrt{3}\).Vertices are at \((0, -4 + \sqrt{3})\) and \((0, -4 - \sqrt{3})\).

Key Concepts

Standard Form Equation

The standard form of a hyperbola's equation is essential when dealing with this type of conic section. It's a specific structure that provides clarity on the hyperbola's orientation and dimensions.

• The general standard form for a hyperbola is: \[ \frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1 \] or \[ \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1 \].
• The choice between these forms depends on whether the transverse axis is parallel to the y-axis or the x-axis.
• For a hyperbola, the right side of the equation is always 1. This differentiates it from ellipses, which also have similar forms but differ in structure.
• When an equation of a hyperbola is given, it's common to first rearrange and simplify it before dividing by the constant to achieve the standard form.

In the exercise provided, after completing the square and simplification, the hyperbola's equation became \[ \frac{(y+4)^2}{3} - x^2 = 1 \], illustrating that the transverse axis is parallel to the y-axis.

Vertices of Hyperbola

Vertices of a hyperbola are specific points where the curve intersects its transverse axis. Recognizing them helps to better understand the shape and dimensions of the hyperbola.

• From the standard form \[ \frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1 \], the vertices are found \(a\) units away from the center along the transverse axis.
• In this problem, the equation \[ \frac{(y+4)^2}{3} - x^2 = 1 \] indicates that the center is at \( (0, -4) \) and the transverse axis is vertical since the y-term comes first.
• Since \( a^2 = 3 \), we have \( a = \sqrt{3} \).
• Thus, the vertices are located at \( (0, -4 + \sqrt{3}) \) and \( (0, -4 - \sqrt{3}) \).

Knowing the vertices' location helps sketch the hyperbola's shape and indicates the extent of its "opening."

Completing the Square

Completing the square is a fundamental technique used to transform quadratic expressions into a perfect square trinomial. This conversion helps in simplifying expressions to either solve or rearrange equations.

• This method is particularly handy when dealing with conic sections like circles, ellipses, and hyperbolas.
• For the quadratic expression \( y^2 + 8y \), the procedure involves taking half of the coefficient of \( y \) (which is 8), squaring it (resulting in 16), and then adding and subtracting this value in the expression: \( y^2 + 8y + 16 - 16 \).
• The expression can then be rewritten as a square: \( (y + 4)^2 - 16 \).
• This makes it possible to easily manipulate the quadratic into a form that relates to the conic section in question, which is crucial for rewriting equations into their standard forms.

By mastering this technique, you'll have a reliable tool for transforming and understanding the equations of various shapes.