Identifying Hyperbola Equations With Center At (2 1)
Hey guys! Today, we're diving into the fascinating world of hyperbolas, specifically focusing on how to identify those whose centers are located at the point (2, 1). We'll be working through several equations, using a bit of algebraic magic to transform them into a standard form that reveals the center. So, buckle up and let's get started!
Understanding Hyperbolas and Their Standard Form
Before we jump into the equations, let's do a quick review of hyperbolas. A hyperbola is a type of conic section, a curve formed by the intersection of a plane and a double cone. Think of it as two mirrored parabolas opening away from each other. The key characteristics of a hyperbola include its center, vertices, foci, and asymptotes. The center, as the name suggests, is the midpoint of the hyperbola. The standard form equation of a hyperbola centered at (h, k) is crucial for identifying these characteristics. There are two forms, depending on whether the hyperbola opens horizontally or vertically:
- Horizontal Hyperbola: ((x - h)^2 / a^2) - ((y - k)^2 / b^2) = 1
- Vertical Hyperbola: ((y - k)^2 / a^2) - ((x - h)^2 / b^2) = 1
In both cases, (h, k) represents the center of the hyperbola. The values of 'a' and 'b' determine the shape and size of the hyperbola, and they are related to the distances from the center to the vertices and co-vertices, respectively. To find hyperbolas centered at (2, 1), we need to manipulate the given equations into one of these standard forms and see if the (h, k) values match our target center. This involves completing the square, a powerful technique that transforms quadratic expressions into perfect squares, making it easier to identify the center and other key features of the hyperbola. Completing the square allows us to rewrite the given equations in a form that directly reveals the center's coordinates. By carefully manipulating the x and y terms, we can reshape the equation to match the standard form, making the identification process straightforward. It's like decoding a secret message, where the standard form is the key to unlocking the hyperbola's center. Remember, the standard form is our roadmap, guiding us through the algebraic transformations needed to pinpoint the center's location. Mastering this process not only helps us solve this particular problem but also lays a strong foundation for understanding and analyzing various conic sections, making us true hyperbola detectives!
Analyzing the Equations
Now, let's tackle the equations one by one. Our goal is to rewrite each equation in the standard form we discussed earlier. We'll be focusing on completing the square for both the x and y terms. This will allow us to clearly identify the center (h, k) of each hyperbola.
Equation 1: -4x^2 + y^2 + 8x - 4y - 4 = 0
First, let's rearrange the terms and group the x and y terms together:
(-4x^2 + 8x) + (y^2 - 4y) = 4
Next, we factor out the coefficients of the squared terms:
-4(x^2 - 2x) + (y^2 - 4y) = 4
Now, we complete the square for both the x and y expressions. Remember, to complete the square, we take half of the coefficient of the linear term, square it, and add it inside the parentheses. Since we factored out coefficients, we need to adjust the values added to the right side of the equation. For the x terms, half of -2 is -1, and (-1)^2 is 1. So we add 1 inside the parentheses, which means we're actually subtracting 4 from the left side (due to the -4 factored out), so we subtract 4 from the right side as well. For the y terms, half of -4 is -2, and (-2)^2 is 4. We add 4 inside the parentheses, so we add 4 to the right side.
-4(x^2 - 2x + 1) + (y^2 - 4y + 4) = 4 - 4 + 4
Now, we can rewrite the expressions as perfect squares:
-4(x - 1)^2 + (y - 2)^2 = 4
Divide both sides by 4 to get the equation in standard form:
-((x - 1)^2 / 1) + ((y - 2)^2 / 4) = 1
((y - 2)^2 / 4) - ((x - 1)^2 / 1) = 1
From this standard form, we can see that the center of this hyperbola is (1, 2). This center does not match our target of (2, 1), so this equation is not a match.
Equation 2: 4x^2 - y^2 - 16x + 2y + 13 = 0
Let's follow the same steps as before. First, rearrange and group terms:
(4x^2 - 16x) + (-y^2 + 2y) = -13
Factor out the coefficients:
4(x^2 - 4x) - (y^2 - 2y) = -13
Complete the square. For the x terms, half of -4 is -2, and (-2)^2 is 4. We add 4 inside the parentheses, which means we're adding 16 to the left side (due to the 4 factored out), so we add 16 to the right side as well. For the y terms, half of -2 is -1, and (-1)^2 is 1. We add 1 inside the parentheses, but we're actually subtracting 1 from the left side (due to the -1 factored out), so we subtract 1 from the right side.
4(x^2 - 4x + 4) - (y^2 - 2y + 1) = -13 + 16 - 1
Rewrite as perfect squares:
4(x - 2)^2 - (y - 1)^2 = 2
Divide both sides by 2:
(2(x - 2)^2) - ((y - 1)^2 / 2) = 1
This is the standard form of a hyperbola. The center of this hyperbola is clearly (2, 1). This equation matches our target center!
Equation 3: x^2 - 4y^2 - 4x + 8y - 4 = 0
Rearrange and group terms:
(x^2 - 4x) + (-4y^2 + 8y) = 4
Factor out coefficients:
(x^2 - 4x) - 4(y^2 - 2y) = 4
Complete the square. For the x terms, half of -4 is -2, and (-2)^2 is 4. We add 4 inside the parentheses, so we add 4 to the right side. For the y terms, half of -2 is -1, and (-1)^2 is 1. We add 1 inside the parentheses, but we're actually subtracting 4 from the left side (due to the -4 factored out), so we subtract 4 from the right side.
(x^2 - 4x + 4) - 4(y^2 - 2y + 1) = 4 + 4 - 4
Rewrite as perfect squares:
(x - 2)^2 - 4(y - 1)^2 = 4
Divide both sides by 4:
((x - 2)^2 / 4) - (y - 1)^2 = 1
The center of this hyperbola is (2, 1)! This equation matches our target center.
Equation 4: 16x^2 + 9y^2 + 64x - 18y - 121 = 0
Rearrange and group terms:
(16x^2 + 64x) + (9y^2 - 18y) = 121
Factor out coefficients:
16(x^2 + 4x) + 9(y^2 - 2y) = 121
Complete the square. For the x terms, half of 4 is 2, and 2^2 is 4. We add 4 inside the parentheses, which means we're adding 64 to the left side (due to the 16 factored out), so we add 64 to the right side as well. For the y terms, half of -2 is -1, and (-1)^2 is 1. We add 1 inside the parentheses, which means we're adding 9 to the left side (due to the 9 factored out), so we add 9 to the right side.
16(x^2 + 4x + 4) + 9(y^2 - 2y + 1) = 121 + 64 + 9
Rewrite as perfect squares:
16(x + 2)^2 + 9(y - 1)^2 = 194
Notice that this equation has a '+' sign between the squared terms, which indicates that this is an ellipse, not a hyperbola. Therefore, we can stop here; this equation doesn't fit our criteria. Ellipses, with their characteristic closed curves, differ significantly from hyperbolas, which exhibit open, diverging branches. The presence of addition between the squared terms is a clear indicator of an ellipse, making it easy to distinguish from a hyperbola's subtraction-based equation. Understanding this fundamental difference is key to accurately classifying conic sections. In the world of conic sections, each shape has its own unique algebraic signature, and recognizing these signatures allows us to quickly and efficiently identify the type of curve we're dealing with. So, while this equation might look similar at first glance, the '+' sign tells us it's an ellipse, and we can confidently move on in our search for hyperbolas.
Equation 5: 16x^2 - 9y^2 - 64x - 18y + 91 = 0
Rearrange and group terms:
(16x^2 - 64x) + (-9y^2 - 18y) = -91
Factor out coefficients:
16(x^2 - 4x) - 9(y^2 + 2y) = -91
Complete the square. For the x terms, half of -4 is -2, and (-2)^2 is 4. We add 4 inside the parentheses, which means we're adding 64 to the left side (due to the 16 factored out), so we add 64 to the right side as well. For the y terms, half of 2 is 1, and (1)^2 is 1. We add 1 inside the parentheses, but we're actually subtracting 9 from the left side (due to the -9 factored out), so we subtract 9 from the right side.
16(x^2 - 4x + 4) - 9(y^2 + 2y + 1) = -91 + 64 - 9
Rewrite as perfect squares:
16(x - 2)^2 - 9(y + 1)^2 = -36
Divide both sides by -36:
-((x - 2)^2 / (9/4)) + ((y + 1)^2 / 4) = 1
((y + 1)^2 / 4) - ((x - 2)^2 / (9/4)) = 1
From the standard form, we can see that the center of this hyperbola is (2, -1). This center does not match our target center of (2, 1).
Final Answer
After analyzing all the equations, we found that only two hyperbolas have their centers at (2, 1):
- 4x^2 - y^2 - 16x + 2y + 13 = 0
- x^2 - 4y^2 - 4x + 8y - 4 = 0
So there you have it! By using the power of completing the square and a solid understanding of the standard form of a hyperbola, we were able to successfully identify the equations with centers at the specified point. Keep practicing, and you'll become a hyperbola master in no time!
