Nearest Points On 5x² - 6xy + 5y² = 4: A Calculus Solution
Hey guys! Ever wondered how to find the closest points on a curve to a specific location, like the origin? It's a classic problem in calculus and optimization, and today we're diving into a fun one: finding the points on the curve 5x² - 6xy + 5y² = 4 that are nearest to the origin. This problem combines concepts from calculus, conic sections, and optimization techniques, making it a fantastic exercise for anyone looking to sharpen their math skills. So, let's break it down step by step, shall we?
Understanding the Problem
Before we jump into the calculations, let's make sure we fully understand the problem. We're given the equation 5x² - 6xy + 5y² = 4, which represents a conic section. Specifically, it's an ellipse, but we'll get into that a bit later. Our goal is to find the points (x, y) on this ellipse that are closest to the origin (0, 0). Think of it like this: we have an oval shape, and we want to find the spots on that oval that are the shortest distance away from the center.
Visualizing the Curve
It can be super helpful to visualize what we're dealing with. Imagine plotting the equation 5x² - 6xy + 5y² = 4 on a graph. You'd see an ellipse centered around the origin, but it's rotated a bit. The rotation comes from the -6xy term. If that term wasn't there, the ellipse would be aligned with the x and y axes, making things a bit simpler. But hey, where's the fun in simple, right?
Defining Distance
To find the points nearest the origin, we need a way to measure distance. Remember the distance formula from way back in geometry class? The distance, d, between any point (x, y) and the origin (0, 0) is given by:
d = √(x² + y²)
Our goal is to minimize this distance, d. But here's a cool trick: instead of minimizing d, we can minimize d², which is just x² + y². Minimizing the square of the distance is easier because it gets rid of the square root, and it doesn't change the location of the minimum points. So, from now on, we'll focus on minimizing f(x, y) = x² + y².
Setting Up the Optimization Problem
Okay, we've got our function to minimize, f(x, y) = x² + y². This is called our objective function. But we can't just pick any values for x and y. They have to satisfy the equation of our ellipse, 5x² - 6xy + 5y² = 4. This equation is our constraint. It limits the possible values of x and y to only those that lie on the curve.
The Method of Lagrange Multipliers
Now, how do we handle this constraint? This is where the Method of Lagrange Multipliers comes to the rescue! This method is a powerful tool for solving optimization problems with constraints. It's like a mathematical magic wand that helps us find the minimum or maximum of a function while staying within the boundaries set by our constraint.
The basic idea behind Lagrange multipliers is to introduce a new variable, usually denoted by λ (lambda), and form a new function called the Lagrangian. The Lagrangian combines our objective function and our constraint. It's defined as:
L(x, y, λ) = f(x, y) - λg(x, y)
where f(x, y) is our objective function (x² + y²) and g(x, y) is the constraint equation rearranged to equal zero. In our case, g(x, y) = 5x² - 6xy + 5y² - 4 = 0. So, our Lagrangian becomes:
L(x, y, λ) = x² + y² - λ(5x² - 6xy + 5y² - 4)
Finding Critical Points
The next step is crucial. We need to find the critical points of the Lagrangian. These are the points where the partial derivatives of L with respect to x, y, and λ are all equal to zero. Think of it like finding the flat spots on a multi-dimensional surface. These flat spots are potential locations for minimums, maximums, or saddle points.
So, let's calculate those partial derivatives:
- ∂L/∂x = 2x - λ(10x - 6y) = 0
- ∂L/∂y = 2y - λ(-6x + 10y) = 0
- ∂L/∂λ = -(5x² - 6xy + 5y² - 4) = 0 (This is just our original constraint equation)
Now we have a system of three equations with three unknowns (x, y, and λ). This might look intimidating, but don't worry, we'll tackle it systematically.
Solving the System of Equations
Alright, we've got our system of equations:
- 2x - λ(10x - 6y) = 0
- 2y - λ(-6x + 10y) = 0
- 5x² - 6xy + 5y² = 4
Our mission is to solve for x, y, and λ. This can be a bit of an algebraic workout, but we'll get there!
Simplifying the Equations
Let's start by simplifying equations 1 and 2. We can rewrite them as:
- 2x = λ(10x - 6y)
- 2y = λ(-6x + 10y)
Now, let's divide both sides of equation 1 by 2x (assuming x ≠ 0) and both sides of equation 2 by 2y (assuming y ≠ 0). This gives us:
- 1 = λ(5 - 3y/x)
- 1 = λ(-3x/y + 5)
Notice anything cool? Both equations now have λ on one side and a similar expression on the other! This means we can set the expressions in the parentheses equal to each other:
5 - 3y/x = -3x/y + 5
Finding the Relationship Between x and y
Let's simplify this equation further. Subtract 5 from both sides and multiply both sides by xy (again, assuming x and y are not zero):
-3y² = -3x²
Divide both sides by -3:
y² = x²
Take the square root of both sides:
y = ±x
This is a crucial result! It tells us that the points nearest the origin must lie on the lines y = x or y = -x. These lines pass through the origin and divide the coordinate plane into four quadrants. This makes sense intuitively, right? The points closest to the origin are likely to be along lines of symmetry.
Substituting into the Constraint Equation
Now that we have a relationship between x and y, we can substitute it back into our constraint equation, 5x² - 6xy + 5y² = 4, to find the specific points. Let's consider the two cases:
Case 1: y = x
Substitute y = x into the constraint equation:
5x² - 6x(x) + 5(x)² = 4
Simplify:
5x² - 6x² + 5x² = 4
4x² = 4
x² = 1
x = ±1
Since y = x, the corresponding y values are also ±1. So, we have two potential points: (1, 1) and (-1, -1).
Case 2: y = -x
Substitute y = -x into the constraint equation:
5x² - 6x(-x) + 5(-x)² = 4
Simplify:
5x² + 6x² + 5x² = 4
16x² = 4
x² = 1/4
x = ±1/2
Since y = -x, the corresponding y values are ∓1/2. So, we have two more potential points: (1/2, -1/2) and (-1/2, 1/2).
Determining the Nearest Points
We've found four candidate points: (1, 1), (-1, -1), (1/2, -1/2), and (-1/2, 1/2). But which ones are actually closest to the origin? To answer this, we need to calculate the distance from each point to the origin. Remember, we're minimizing d² = x² + y².
- For (1, 1) and (-1, -1): d² = 1² + 1² = 2
- For (1/2, -1/2) and (-1/2, 1/2): d² = (1/2)² + (-1/2)² = 1/4 + 1/4 = 1/2
Aha! The points (1/2, -1/2) and (-1/2, 1/2) have the smallest value of d², which means they are the closest points to the origin on the curve 5x² - 6xy + 5y² = 4.
Final Answer
So, there you have it! The points on the curve 5x² - 6xy + 5y² = 4 that are nearest the origin are (1/2, -1/2) and (-1/2, 1/2). We solved this problem by understanding the geometry, setting up an optimization problem with constraints, using the Method of Lagrange Multipliers, solving a system of equations, and finally, comparing distances to find the minimum. What a journey!
This problem is a great example of how different areas of math come together to solve real-world (or, well, abstract-world) problems. Keep practicing, keep exploring, and who knows what mathematical mysteries you'll unravel next!