Problem 80 - A Graphical Approach to College Algebra

Question

Write the equation in standard form for a hyperbola centered at ( \(h, k\) ). Identify the center and vertices. \(4 x^{2}+16 x-9 y^{2}+18 y=29\)

Step-by-Step Solution

Answer

Center: (-2, 1); Vertices: (-5, 1) and (1, 1).

Step 1: Rearrange Terms

Start by rearranging the terms so that the variables with x are together and the variables with y are together. Write it as:\[ 4x^2 + 16x - 9y^2 + 18y = 29 \]

Step 2: Group Similar Terms

Group the x-terms and y-terms together:\[ (4x^2 + 16x) - (9y^2 - 18y) = 29 \]

Step 3: Complete the Square for x

To complete the square for the x terms, factor out the coefficient of the quadratic term:\[ 4(x^2 + 4x) \]Complete the square inside the parentheses:\[ x^2 + 4x \Rightarrow (x+2)^2 - 4 \]Thus:\[ 4((x+2)^2 - 4) = 4(x+2)^2 - 16 \]

Step 4: Complete the Square for y

Similarly, for the y terms, factor out the coefficient:\[ -9(y^2 - 2y) \]Complete the square for y:\[ y^2 - 2y \Rightarrow (y-1)^2 - 1 \]Thus:\[ -9((y-1)^2 - 1) = -9(y-1)^2 + 9 \]

Step 5: Substitute and Simplify

Substitute the completed squares back into the equation:\[ 4(x+2)^2 - 16 - 9(y-1)^2 + 9 = 29 \]Combine the constants on the right side:\[ 4(x+2)^2 - 9(y-1)^2 = 36 \]

Step 6: Divide Through by 36 to Standardize Equation

Divide every term by 36 to convert the equation into standard form:\[ \frac{4(x+2)^2}{36} - \frac{9(y-1)^2}{36} = \frac{36}{36} \]Simplifying:\[ \frac{(x+2)^2}{9} - \frac{(y-1)^2}{4} = 1 \]

Step 7: Identify the Center and Vertices

The standard form of the hyperbola is now given by:\[ \frac{(x+2)^2}{9} - \frac{(y-1)^2}{4} = 1 \]From this, the center of the hyperbola is (\(-2, 1\)). The vertices are located \(3\) units left and right from the center along the x-axis, giving the vertices' coordinates as \((-5, 1)\) and \((1, 1)\).

Key Concepts

Completing the Square for Hyperbolas

Completing the square is a method used to convert quadratic expressions into a more workable form. This technique is especially useful when working with hyperbolas, as it aids in transforming the equation into its standard form. In the given exercise, once we've grouped the terms, we focus on each variable separately.

• Start by isolating the quadratic term of each variable. For instance, with the x terms: from \(4x^2 + 16x\), factor out the 4 to get \(x^2 + 4x\).
• Next, complete the square by taking half of the coefficient of \(x\) (which is 4), squaring it to get 4, and then rewriting the expression as \((x+2)^2 - 4\).
• Similarly, for the y terms \(-9(y^2 - 2y)\), you factor out \(-9\), leaving you with \(y^2 - 2y\).
• Complete the square by adding and subtracting 1 inside the bracket, giving you \((y-1)^2 - 1\).

This manipulation of terms ultimately allows for the final rewriting of the equation in a recognizable standard form for a hyperbola.

Identifying the Center of a Hyperbola

To find the center of a hyperbola, it is crucial to rewrite the equation in its standard form. The center of the hyperbola \[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \] is represented by the point \((h, k)\).
By completing the square, we rewrite the equation in the given exercise as \[ \frac{(x+2)^2}{9} - \frac{(y-1)^2}{4} = 1 \].

• The center is derived from the transformations on \(x\) and \(y\). Here, it is \((h, k) = (-2, 1)\).
• This is due to the fact we have \((x+2)^2\) and \((y-1)^2\), indicating a shift of the hyperbola.

The center provides a point of reference for other features like vertices and asymptotes, and knowing it helps us understand the positioning of the hyperbola in the coordinate plane.

Determining Vertices of a Hyperbola

Vertices of a hyperbola are key points that lie on the transverse axis, providing indications of the hyperbola’s width. Once the equation is expressed in standard form, identifying the vertices becomes straightforward.

• In the standard form \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\), the vertices are directly affected by \(a\), the distance from the center to a vertex along the x-axis in this case (since our major axis lies along x).
• Given our equation \(\frac{(x+2)^2}{9} - \frac{(y-1)^2}{4} = 1\), we see that \(a^2 = 9\), therefore \(a = 3\).
• As such, the vertices are \(3\) units to the left and right of the center, \((-2, 1)\), along the x-axis, resulting in vertices at \((-5, 1)\) and \((1, 1)\).

Understanding the placement of vertices assists in sketching the hyperbola and is essential for further calculations involving its asymptotes and foci.