Hyperbola 4x² - Y² = 16: Calculating Focal Distance

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Hey everyone! Today, we're diving into the fascinating world of hyperbolas, specifically focusing on how to calculate the focal distance of a hyperbola. We'll take the equation 4x² - y² = 16 as our example. This might sound intimidating at first, but don't worry, we'll break it down step by step so that even if you're just starting with conic sections, you'll be able to follow along. We'll cover the basics of hyperbolas, the standard form equation, and then get into the nitty-gritty of calculating the focal distance. So, grab your calculators, and let's get started!

Understanding Hyperbolas

Before we jump into the calculations, let's make sure we're all on the same page about what a hyperbola actually is. In simple terms, 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. Now, hyperbolas have some key features that we need to understand to calculate the focal distance. These include:

  • Foci (plural of focus): These are two fixed points inside the hyperbola that define its shape. The focal distance is directly related to the position of these foci.
  • Vertices: These are the points where the hyperbola intersects its major axis (the axis that passes through the foci).
  • Center: The midpoint between the two foci (and also the midpoint between the vertices).
  • Asymptotes: These are lines that the hyperbola approaches as it extends towards infinity. They help define the “opening” of the hyperbola.

Understanding these basic elements is crucial because they all play a role in determining the focal distance. The distance between the center and each focus is what we're aiming to calculate. To get there, we need to look at the standard form equation of a hyperbola.

Standard Form Equation of a Hyperbola

The standard form equation is our key to unlocking the focal distance. Guys, remember, the standard form gives us the essential information we need – the values of 'a' and 'b,' which are directly related to the semi-major axis and semi-minor axis of the hyperbola. There are two forms, depending on whether the hyperbola opens horizontally or vertically:

  • Horizontal Hyperbola: (x²/a²) - (y²/b²) = 1
  • Vertical Hyperbola: (y²/a²) - (x²/b²) = 1

In our case, we have the equation 4x² - y² = 16. The first step is to get this equation into standard form. We do this by dividing both sides by 16:

(4x²/16) - (y²/16) = 1

Simplifying, we get:

(x²/4) - (y²/16) = 1

Now, we can clearly see that this is a horizontal hyperbola (because the x² term is positive) and that a² = 4 and b² = 16. This is a huge step forward! We've extracted the key parameters from the equation. Now, we need to use these values to calculate the focal distance. Remember, ‘a’ represents the distance from the center to each vertex along the major axis, and ‘b’ is related to the distance along the minor axis. These values, along with a crucial formula, will lead us to our goal.

Calculating the Focal Distance

Okay, now for the exciting part: calculating the focal distance! The focal distance, often denoted as 'c', is the distance from the center of the hyperbola to each focus. The relationship between 'a', 'b', and 'c' in a hyperbola is given by the following equation:

c² = a² + b²

This equation is super important, so make sure you've got it locked in! It's similar to the Pythagorean theorem, but with a plus sign instead of a minus sign. Remember, we already found that a² = 4 and b² = 16 from our standard form equation. Now, we can plug these values into the equation:

c² = 4 + 16

c² = 20

To find 'c', we take the square root of both sides:

c = √20

Simplifying the square root, we get:

c = 2√5

So, the focal distance 'c' is 2√5. This means that each focus is located 2√5 units away from the center of the hyperbola along the major axis. We've done it! We've successfully calculated the focal distance. But let's take it a step further and discuss what this result actually means and how it relates to the graph of the hyperbola.

Understanding the Result and Graphing the Hyperbola

Great job, guys! We've calculated that the focal distance (c) for the hyperbola 4x² - y² = 16 is 2√5. But what does this number actually tell us? And how does it help us visualize the hyperbola? The focal distance, as we've discussed, is the distance from the center of the hyperbola to each of its foci. Since our hyperbola is in the form (x²/4) - (y²/16) = 1, we know it opens horizontally. The center of this hyperbola is at the origin (0, 0). Because c = 2√5, the foci are located at the points (2√5, 0) and (-2√5, 0).

Knowing the foci is crucial for accurately graphing the hyperbola. We also know that a² = 4, so a = 2. This means the vertices (the points where the hyperbola intersects its major axis) are located at (2, 0) and (-2, 0). We also know that b² = 16, so b = 4. The value of 'b' helps us determine the asymptotes of the hyperbola. The asymptotes are lines that the hyperbola approaches as it extends infinitely. For a horizontal hyperbola centered at the origin, the equations of the asymptotes are y = (b/a)x and y = -(b/a)x. In our case, this translates to y = (4/2)x and y = -(4/2)x, which simplifies to y = 2x and y = -2x. By plotting the center, vertices, foci, and asymptotes, we can accurately sketch the hyperbola. The hyperbola will curve away from the center, passing through the vertices and approaching the asymptotes. Visualizing the hyperbola in this way helps solidify our understanding of how the focal distance, vertices, and asymptotes all work together to define the shape of the hyperbola. Remember, the larger the focal distance (c) relative to 'a', the