Problem 59

Question

For the following exercises, express the equation for the hyperbola as two functions, with $y$ as a function of $x .$ Express as simply as possible. Use a graphing calculator to sketch the graph of the two functions on the same axes. $$ -4 x^{2}-16 x+y^{2}-2 y-19=0 $$

Step-by-Step Solution

Verified

Answer

The functions are $y = 1 \pm \sqrt{4 + 4(x+2)^2}$.

1Step 1: Rearrange the Equation

The given equation is $ -4x^2 - 16x + y^2 - 2y - 19 = 0 $. Our first task is to rearrange this equation to separate the $x$ terms and the $y$ terms. Begin by moving $19$ to the other side of the equation: $ -4x^2 - 16x + y^2 - 2y = 19 $.

2Step 2: Complete the Square for x

We will complete the square for the $x$ terms. Start with $-4x^2 - 16x$. Factor out $-4$ from these terms: $-4(x^2 + 4x)$. To complete the square inside the parenthesis, take half of $4$, square it to get $4$, and add inside the square: $-4((x+2)^2 - 4)$. This simplifies to $-4(x+2)^2 + 16$.

3Step 3: Complete the Square for y

Now complete the square for the $y$ terms in $y^2 - 2y$. Take half of $-2$, square it to get $1$, and complete the square: $(y-1)^2 - 1$.

4Step 4: Combine Completed Squares

Insert these into the equation: $-4(x+2)^2 + 16 + (y-1)^2 - 1 = 19$. Simplifying gives $-4(x+2)^2 + (y-1)^2 = 4$.

5Step 5: Express y as a Function of x

Rearrange the equation for $y$: $(y-1)^2 = 4 + 4(x+2)^2$. Solve for $y-1$ which gives $y-1 = \pm \sqrt{4 + 4(x+2)^2}$. Hence, $y = 1 \pm \sqrt{4 + 4(x+2)^2}$.

6Step 6: Graph the Functions

The two functions are $y = 1 + \sqrt{4 + 4(x+2)^2}$ and $y = 1 - \sqrt{4 + 4(x+2)^2}$. Use a graphing calculator or plotting tool to graph these functions on the same set of axes to visualize the hyperbola.

Key Concepts

Equation RearrangementCompleting the SquareGraphing FunctionsCoordinate Geometry

Equation Rearrangement

When dealing with hyperbolas, the first task often involves manipulating the given equation into a more manageable form. This is known as equation rearrangement. The initial expression for our hyperbola is:\[-4x^2 - 16x + y^2 - 2y - 19 = 0\]Begin by isolating the constant term by moving it to the other side of the equation. This action gives:\[-4x^2 - 16x + y^2 - 2y = 19\]Equation rearrangement is crucial because it sets the stage for completing the square, which further helps in simplifying functions for finding their graphical representation. Pay attention to grouping terms involving the same variable, as this will aid in subsequent steps.

Completing the Square

Completing the square is a technique employed to transform quadratic expressions into a perfect square trinomial, which simplifies equations involving hyperbolas. In our case, we need to complete the square for both the $x$ and $y$ terms separately.

For the $x$ Terms:

Work with the expression $-4x^2 - 16x$. Factor out $-4$:\[-4(x^2 + 4x)\]Now take half of the $4$ in $x^2 + 4x$, square it, and add to form a perfect square trinomial:\[-4((x+2)^2 - 4)\]This simplifies further to:\[-4(x+2)^2 + 16\]

For the $y$ Terms:

The process is similar. Start by looking at $y^2 - 2y$. To complete the square, take half of $-2$, square it, which provides $1$:\[(y-1)^2 - 1\]The completion of squares not only simplifies the equation but also reveals displaced centers of the hyperbola, pivotal for future calculations.

Graphing Functions

Once we've completed the squares for both $x$ and $y$ terms and simplified the equation, graphing becomes the next logical step. Graphing allows visualization of the hyperbola and aids in understanding its geometric properties.The equation turns into a recognizable form:\[-4(x+2)^2 + (y-1)^2 = 4\]When isolating $y$, obtain two functions:1. $y = 1 + \sqrt{4 + 4(x+2)^2}$2. $y = 1 - \sqrt{4 + 4(x+2)^2}$Plot these expressions using graphing tools like graphing calculators or software. When graphing, note each branch of the hyperbola. They should mirror each other, demonstrating symmetry about both axes. Recognize how changes in $x$ affect $y$, illustrating the nature of the hyperbola.

Coordinate Geometry

Coordinate geometry helps us make sense of the hyperbola's structure on the Cartesian plane. Hyperbolas are defined by their distinctive open curve shape with two branches, oriented away from each other.Key points when examining a hyperbola through coordinate geometry:

Center Position: Defined by the completed square process, in this exercise the center is at $(-2,1)$.
Axes: The orientation of the hyperbola's axes depends on the coefficients and terms derived from rearranging and completing squares.
Vertices and Foci: These points determine the extent and sharpness of the curve, calculated from the equation.

Understanding coordinate geometry involves recognizing these elements from the equation to predict and describe the graph's behavior. It shows the practical side of hyperbolas in math, illustrating its structure within a coordinate system.

Problem 59