Finding Foci: Hyperbola Equation 9x² - 4y² - 18x - 16y - 43 = 0

Hey guys! Today, we're diving into the fascinating world of conic sections, specifically hyperbolas! We're going to tackle a common problem: finding the foci of a hyperbola given its general equation. It might seem a bit daunting at first, but don't worry, we'll break it down step-by-step so you can master this skill. Our specific example is the hyperbola defined by the equation 9x² - 4y² - 18x - 16y - 43 = 0. So, grab your pencils and let's get started!

Understanding Hyperbolas and Their Foci

Before we jump into the calculations, let's make sure we're all on the same page about what a hyperbola is and what we mean by its foci. A hyperbola is a type of conic section, which basically means it's a curve formed by the intersection of a plane and a double cone. Think of two identical cones placed tip-to-tip, and then slice them with a plane – the resulting shape is a hyperbola. Hyperbolas have some key features, including two branches that open away from each other, a center, two vertices (the points where the branches are closest to each other), and, most importantly for our task today, two foci.

The foci (plural of focus) are two fixed points inside the hyperbola. They play a crucial role in defining the shape of the hyperbola. In fact, a hyperbola can be defined as the set of all points where the difference of the distances to the two foci is constant. This is different from an ellipse, where the sum of the distances to the foci is constant. Understanding this fundamental property is key to appreciating the geometry of hyperbolas. The foci are always located on the transverse axis, which is the line that passes through the vertices and the center of the hyperbola. The distance between the center and each focus is denoted by 'c', which we'll calculate later. Visualizing these elements helps in solving problems related to hyperbolas. So, remember, we're looking for these two special points that dictate the hyperbola's unique shape!

Step 1: Convert the General Equation to Standard Form

The first thing we need to do is transform the given general equation, 9x² - 4y² - 18x - 16y - 43 = 0, into its standard form. The standard form of a hyperbola equation makes it much easier to identify the key parameters we need, such as the center, the lengths of the semi-major and semi-minor axes (a and b), and ultimately, the distance to the foci (c). The standard form equation depends on whether the hyperbola opens horizontally or vertically.

To get there, we'll use a technique called completing the square. This is a handy algebraic trick that allows us to rewrite quadratic expressions in a more manageable form. Basically, we'll group the x terms and the y terms together, complete the square for each group, and then rearrange the equation to match the standard form. Let's start by grouping the x and y terms: (9x² - 18x) - (4y² + 16y) = 43. Notice we factored out the negative sign from the y terms. Now, we need to complete the square for the expressions inside the parentheses. For the x terms, we factor out the 9: 9(x² - 2x). To complete the square, we take half of the coefficient of the x term (-2), square it ((-1)² = 1), and add it inside the parentheses. So we have 9(x² - 2x + 1). But remember, we've actually added 9 * 1 = 9 to the left side of the equation, so we need to add 9 to the right side as well. For the y terms, we factor out the 4: -4(y² + 4y). Half of the coefficient of the y term (4) is 2, and squaring it gives us 4. So we add 4 inside the parentheses: -4(y² + 4y + 4). This means we've subtracted 4 * 4 = 16 from the left side, so we need to subtract 16 from the right side as well. Now our equation looks like this: 9(x² - 2x + 1) - 4(y² + 4y + 4) = 43 + 9 - 16. We can now rewrite the expressions in parentheses as perfect squares: 9(x - 1)² - 4(y + 2)² = 36. Finally, we divide both sides by 36 to get the equation in standard form: (x - 1)² / 4 - (y + 2)² / 9 = 1. This is the standard form of our hyperbola equation, and it unlocks a wealth of information!

Step 2: Identify the Center, a, and b

Now that we have the standard form equation, (x - 1)² / 4 - (y + 2)² / 9 = 1, we can easily identify the key parameters that define our hyperbola. These parameters are essential for finding the foci. Let's break down what each part of the equation tells us. The standard form equation of a hyperbola that opens horizontally is (x - h)² / a² - (y - k)² / b² = 1, where (h, k) is the center of the hyperbola, 'a' is the distance from the center to each vertex along the transverse axis, and 'b' is related to the distance along the conjugate axis. If the equation were (y - k)² / a² - (x - h)² / b² = 1, the hyperbola would open vertically. Comparing our equation to the standard form, we can see that:

• The center of the hyperbola is (h, k) = (1, -2). This is the midpoint of the hyperbola, and it's our starting point for finding the foci.
• a² = 4, so a = 2. This tells us that the vertices are 2 units away from the center along the x-axis (since the x² term comes first and is positive, indicating a horizontal hyperbola).
• b² = 9, so b = 3. This value is related to the shape of the hyperbola but doesn't directly give us the distance to the foci.

Identifying these values is a crucial step because they feed into the next calculation: finding 'c', the distance from the center to each focus. With the center, 'a', and 'b' in hand, we're well on our way to locating the foci!

Step 3: Calculate c (Distance from Center to Foci)

With 'a' and 'b' determined, we can now calculate 'c', the distance from the center of the hyperbola to each focus. This is a critical value because it tells us exactly how far to move along the transverse axis from the center to find the foci. The relationship between a, b, and c in a hyperbola is given by the equation c² = a² + b². This equation is similar to the Pythagorean theorem, but it's important to remember that it's specific to hyperbolas (and ellipses have a slightly different relationship). We already know that a² = 4 and b² = 9, so we can plug these values into the equation:

c² = 4 + 9 c² = 13

Taking the square root of both sides, we get c = √13. This is the distance from the center of the hyperbola to each focus. Since √13 is approximately 3.61, we know that the foci are a little more than 3.6 units away from the center along the transverse axis. This value of 'c' is the key to pinpointing the exact locations of the foci.

Step 4: Determine the Coordinates of the Foci

Now that we know the center of the hyperbola is (1, -2) and the distance from the center to each focus is c = √13, we can finally determine the coordinates of the foci. Remember, since the x² term is positive in the standard form equation, the hyperbola opens horizontally. This means the foci lie on the horizontal line that passes through the center. To find the foci, we'll move a distance of 'c' units to the left and right from the center along this line.

The coordinates of the center are (1, -2). So, to find the foci:

• Focus 1: Move √13 units to the right from the center: (1 + √13, -2)
• Focus 2: Move √13 units to the left from the center: (1 - √13, -2)

Therefore, the foci of the hyperbola are (1 + √13, -2) and (1 - √13, -2). We've successfully located the two points that define the shape of our hyperbola! We can also approximate these coordinates using the value of √13 ≈ 3.61:

• Focus 1 ≈ (1 + 3.61, -2) ≈ (4.61, -2)
• Focus 2 ≈ (1 - 3.61, -2) ≈ (-2.61, -2)

These approximate values can be helpful for visualizing the location of the foci on a graph.

Conclusion: You've Found the Foci!

Great job, guys! You've successfully navigated the process of finding the foci of a hyperbola given its general equation. We started by understanding the definition of a hyperbola and its foci, then we converted the general equation to standard form using the completing the square technique. From the standard form, we identified the center, 'a', and 'b', and then calculated 'c', the distance from the center to the foci. Finally, we used the center and the value of 'c' to determine the coordinates of the foci.

The foci of the hyperbola defined by the equation 9x² - 4y² - 18x - 16y - 43 = 0 are (1 + √13, -2) and (1 - √13, -2). Remember, practice makes perfect, so try working through similar problems to solidify your understanding. And don't hesitate to review these steps whenever you encounter a hyperbola challenge. Keep exploring the fascinating world of conic sections!