Finding The Locus Of Point P Where |PA-PB| = 6
Hey guys! Ever stumbled upon a math problem that looks like it's speaking a different language? Today, we're diving deep into one of those, but don't worry, we'll break it down together. We're tackling a locus problem – sounds intimidating, right? But trust me, it's like solving a puzzle. Our mission, should we choose to accept it, is to find the equation of the locus of a point P, where the absolute difference of its distances from two fixed points A and B is a constant. Specifically, we're given |PA - PB| = 6, with A at (0, 2) and B at (0, -2). Buckle up; let's unravel this mathematical mystery!
Understanding the Locus Problem
So, what exactly is a locus in the world of mathematics? Think of it as a path – not just any path, but the trail left behind by a point that moves according to a specific set of rules. In our case, point P is cruising around the coordinate plane, but it's not free to roam wherever it pleases. It's bound by the condition that the absolute difference in its distances from points A and B must always be 6. This constraint dictates P's path, and that path is what we call the locus. To truly grasp the essence of this problem, we need to visualize it. Imagine A and B as fixed anchors, and P as a wandering soul tethered to them by invisible strings. The difference in the lengths of these strings remains constant, carving out a distinct shape as P moves. This shape, my friends, is what we're after – the locus of P. It's not just about finding a single point; it's about mapping out an entire path, an infinite set of points that all satisfy the given condition. Now, let's roll up our sleeves and get into the nitty-gritty of how we're going to find this locus. We'll be using some cool mathematical tools and techniques, but don't sweat it – we'll take it one step at a time. Our journey starts with understanding the distance formula, a trusty companion for any locus-hunting expedition.
Applying the Distance Formula
The distance formula is our trusty tool in this quest. It helps us calculate the distance between two points in a coordinate plane. Remember it? It's like the Pythagorean theorem's cool cousin: √((x₂ - x₁)² + (y₂ - y₁)²). We're going to use this formula to express the distances PA and PB in terms of the coordinates of point P, which we'll call (x, y). So, PA is the distance between P(x, y) and A(0, 2), and PB is the distance between P(x, y) and B(0, -2). Let's plug these coordinates into the distance formula and see what we get. For PA, we have √((x - 0)² + (y - 2)²), which simplifies to √(x² + (y - 2)²). And for PB, we have √((x - 0)² + (y - (-2))²), which simplifies to √(x² + (y + 2)²). Now, we've got expressions for PA and PB in terms of x and y. But remember, our goal isn't just to find these distances; it's to use them in the given condition: |PA - PB| = 6. This equation is the key to unlocking the locus of P. It tells us that the absolute difference between these two distances is always 6. So, we're going to take these expressions for PA and PB, plug them into this equation, and then… well, that's where the real fun begins. We'll need to do some algebraic gymnastics to simplify the equation and get it into a form that we recognize. But don't worry, we'll take it one step at a time. The first step is simply substituting our distance expressions into the given condition. Let's do it!
Setting up the Equation |PA - PB| = 6
Alright, guys, we've got our expressions for PA and PB, and we know that |PA - PB| = 6. It's time to put these pieces together and form our equation. Substituting the distance formulas into our condition, we get |√(x² + (y - 2)²) - √(x² + (y + 2)²) | = 6. Whoa, that looks a bit intimidating, right? Don't worry; we're not going to let this equation scare us. It might seem complex now, but with a bit of algebraic finesse, we can tame it. The absolute value signs mean that we have two scenarios to consider: either √(x² + (y - 2)²) - √(x² + (y + 2)²) = 6 or √(x² + (y - 2)²) - √(x² + (y + 2)²) = -6. Both cases are equally valid, and we need to account for both to fully describe the locus of P. Now, the challenge is to get rid of those square roots and absolute value signs. That's where our algebraic skills come into play. We'll be squaring both sides, rearranging terms, and simplifying until we arrive at an equation that we recognize – hopefully, one that describes a familiar geometric shape. This process might involve a few steps, but each step brings us closer to our goal. So, let's take a deep breath and dive into the algebra. Our first move is to isolate one of the square roots on one side of the equation. This will make the squaring process a bit cleaner. Let's isolate the first square root in both cases and see what happens.
Simplifying the Equation: Squaring and Rearranging
Okay, so we've got our equation with the square roots, and we're ready to start simplifying. The key to taming those square roots is to square both sides of the equation. But before we do that, let's isolate one of the square root terms. This makes the algebra a bit less messy. Let's consider the case where √(x² + (y - 2)²) - √(x² + (y + 2)²) = 6. We can rewrite this as √(x² + (y - 2)²) = 6 + √(x² + (y + 2)²). Now, we square both sides. Remember, when we square a binomial (like 6 + √(x² + (y + 2)²)), we need to use the formula (a + b)² = a² + 2ab + b². So, squaring both sides, we get: x² + (y - 2)² = 36 + 12√(x² + (y + 2)²) + (x² + (y + 2)²). See? We've gotten rid of one square root, but another one popped up in its place. Don't worry; this is all part of the plan. Now, let's expand those squared terms and see if we can simplify things further. We have x² + y² - 4y + 4 = 36 + 12√(x² + (y + 2)²) + x² + y² + 4y + 4. Notice that we have x² and y² terms on both sides, which we can cancel out. We also have a +4 on both sides, which we can cancel too. This leaves us with -4y = 36 + 12√(x² + (y + 2)²) + 4y. Now, let's get all the non-square root terms on one side. We can subtract 4y from both sides to get -8y = 36 + 12√(x² + (y + 2)²). Things are starting to look a bit more manageable, right? We still have that pesky square root, but we're one step closer to getting rid of it. We can divide the entire equation by 4 to simplify it further: -2y = 9 + 3√(x² + (y + 2)²). Now, we're ready for our next move: isolating the remaining square root and squaring both sides again. This might seem like a lot of work, but trust me, it's worth it. We're gradually unraveling the equation and revealing the hidden shape of the locus.
Squaring Again and Identifying the Locus
Alright, guys, we're on the home stretch! We've simplified our equation to -2y = 9 + 3√(x² + (y + 2)²). Now, let's isolate the square root term again. Subtracting 9 from both sides gives us -2y - 9 = 3√(x² + (y + 2)²). Now comes the moment we've been waiting for: squaring both sides again! Remember, we need to be careful when squaring the left side, which is a binomial. We use the formula (a + b)² = a² + 2ab + b². So, squaring both sides, we get: (-2y - 9)² = (3√(x² + (y + 2)²))². This expands to 4y² + 36y + 81 = 9(x² + (y + 2)²). Let's expand the right side as well: 4y² + 36y + 81 = 9(x² + y² + 4y + 4). Distributing the 9, we have 4y² + 36y + 81 = 9x² + 9y² + 36y + 36. Now, let's get all the terms on one side and see if we can simplify. Subtracting 4y², 36y, and 81 from both sides gives us 0 = 9x² + 5y² - 45. Rearranging, we get 9x² + 5y² = 45. Now, this is starting to look familiar! To make it even clearer, let's divide both sides by 45: (9x²/45) + (5y²/45) = 1. Simplifying, we get x²/5 + y²/9 = 1. Bam! We've done it! This, my friends, is the equation of an ellipse. An ellipse is like a stretched-out circle, and its equation has the form x²/a² + y²/b² = 1, where a and b are the semi-major and semi-minor axes. In our case, a² = 5 and b² = 9, so a = √5 and b = 3. This means our ellipse is centered at the origin (0, 0), with its major axis along the y-axis (since b > a). So, the locus of point P is an ellipse! We've taken a complex problem, broken it down step by step, and arrived at a beautiful geometric shape. High five!
Conclusion: The Elliptical Journey
Wow, what a journey! We started with a seemingly complex problem involving distances and absolute values, and we ended up discovering a beautiful ellipse. That's the magic of math, guys! We learned how to find the locus of a point by applying the distance formula, setting up equations, and simplifying them using algebraic techniques. We squared both sides (twice!), rearranged terms, and finally arrived at the equation x²/5 + y²/9 = 1, which represents an ellipse. This problem wasn't just about finding an answer; it was about the process of problem-solving. We learned how to break down a complex problem into smaller, manageable steps. We learned how to use the distance formula and algebraic manipulation to our advantage. And most importantly, we learned that even the most intimidating-looking problems can be solved with patience, persistence, and a little bit of math magic. So, the next time you encounter a locus problem, don't shy away. Remember our journey today, and remember that you have the tools and the knowledge to tackle it. Keep exploring, keep questioning, and keep solving! And who knows, maybe you'll discover another hidden ellipse along the way. Keep up the awesome work, mathletes!