Problem 83 - 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. \(4 y^{2}+32 y-5 x^{2}-10 x+39=0\)

Step-by-Step Solution

Answer

The center is (-1, -4); vertices are \((-1, -4+\sqrt{5})\) and \((-1, -4-\sqrt{5})\).

Step 1: Rearrange Terms

First, rearrange the given equation by grouping the terms with variables and moving the constant to the other side of the equation: \[ 4y^2 + 32y - 5x^2 - 10x = -39 \]

Step 2: Complete the Square

To convert the equation into standard form, complete the square for both the \(y\)-terms and the \(x\)-terms:For the \(y\)-terms:- Factor out the coefficient of \(y^2\):\[ 4(y^2 + 8y) \]- Complete the square by adding and subtracting 16:\[ 4(y^2 + 8y + 16 - 16) = 4((y + 4)^2 - 16) \]- Simplify to:\[ 4(y + 4)^2 - 64 \]For the \(x\)-terms:- Factor out the coefficient of \(x^2\):\[ -5(x^2 + 2x) \]- Complete the square by adding and subtracting 1:\[ -5(x^2 + 2x + 1 - 1) = -5((x + 1)^2 - 1) \]- Simplify to:\[ -5(x + 1)^2 + 5 \]

Step 3: Substitute and Simplify

Substitute the completed square forms back into the equation:\[ 4(y + 4)^2 - 64 - 5(x + 1)^2 + 5 = -39 \]Simplify the equation:\[ 4(y + 4)^2 - 5(x + 1)^2 = -39 + 64 - 5 \]This simplifies to:\[ 4(y + 4)^2 - 5(x + 1)^2 = 20 \]

Step 4: Convert to Standard Form

Divide the entire equation by 20 to express it in standard form:\[ \frac{(y + 4)^2}{5} - \frac{(x + 1)^2}{4} = 1 \]This is the standard form for a hyperbola.

Step 5: Identify Center and Vertices

The standard form of the hyperbola is \( \frac{(y + 4)^2}{5} - \frac{(x + 1)^2}{4} = 1 \), indicating a vertically oriented hyperbola.- Center: \((h, k) = (-1, -4)\)- Vertices (since it's vertical): The vertices are a distance of \(a = \sqrt{5}\) from the center along the y-axis. Thus, the vertices are\[ (-1, -4+\sqrt{5}) \text{ and } (-1, -4-\sqrt{5}) \]

Key Concepts

Completing the Square

Completing the square is a method used to transform quadratic expressions into a perfect square trinomial. This transformation makes it easier to manipulate and solve these equations, especially when dealing with conic sections such as hyperbolas.
To complete the square, you need to find a number that, when added and subtracted to a binomial, turns it into a perfect square trinomial. For instance, in the equation given in the exercise, let's complete the square for the terms with variable y:

• Identify the coefficient of the linear y-term. Here, it's 8 (from 8y).
• Divide this coefficient by 2, giving 4, and square it to find the number to add and subtract: 16.
• Add and subtract this value inside the bracket to maintain balance: \( 4(y^2 + 8y + 16 - 16) \)

The quadratic (y+4)^2 can now be used in further transformations.

Standard Form of a Conic Section

A conic section represents the intersection of a plane with a cone and includes shapes like circles, ellipses, parabolas, and hyperbolas.
The standard form is pivotal as it simplifies understanding and computation of these shapes' properties.
For hyperbolas, the standard form helps us quickly identify the center, axes, and vertices:

• If the hyperbola is oriented vertically, as in our example, the standard form is \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \).
• If it's horizontal, the form reverses the roles of x and y.

The process of rewriting a given equation to standard form involves isolating perfect square trinomial expressions and adjusting the equation accordingly. Through factoring and simplifying, we achieve an equation that is easy to analyze and use.

Center of a Hyperbola

The center of a hyperbola is a key point from which the axes of symmetry and positioning of the curve are derived. In the standard form equation, the center is given by coordinates (h, k).
For the hyperbola in the exercise, the standard form equation is \( \frac{(y + 4)^2}{5} - \frac{(x + 1)^2}{4} = 1 \).
The center of this hyperbola is (-1, -4), obtained by taking the values that revert each square term back to zero in the standard form.

• Analyze the equation components: recognize that the transformation inside the squared terms like (y+4) and (x+1) indicate a shift from the origin.
• The signs are switched when identifying the center, hence, y+4 indicates k = -4, and x+1 indicates h = -1.

These transformations ensure the hyperbola is centered correctly in the coordinate plane.

Vertices of a Hyperbola

Vertices are critical points on the hyperbola, showing the closest and furthest curves from the center along the axes. The standard form offers an easy way to find these.
For a hyperbola, once in the standard form, \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \),
vertices are located \pm a units away from the center along the direction of the major axis.

• In our equation \( \frac{(y + 4)^2}{5} - \frac{(x + 1)^2}{4} = 1 \), we find \( a^2 = 5 \), giving \( a = \sqrt{5} \).
• Since our hyperbola is vertical, the vertices will be along the y-axis from the center (-1, -4).
• Hence, add and subtract \( \sqrt{5} \) from -4 to find the vertices: (-1, -4+\sqrt{5}) and (-1, -4-\sqrt{5}).

Using the knowledge of center and axis, calculating vertices becomes a simple application of the standard form properties.