Nearest Points On 5x² - 6xy + 5y² = 4: A Calculus Solution

Aug 14, 2025 by Kenji Nakamura 59 views

Finding Points Closest to the Origin on the Curve 5x² - 6xy + 5y² = 4

Introduction

Hey everyone! Today, we're diving into a classic calculus problem that combines optimization techniques with conic sections. We're going to figure out how to find the points on the curve defined by the equation 5x² - 6xy + 5y² = 4 that are nearest to the origin (0, 0). This problem is a fantastic example of how calculus can be used to solve real-world geometric questions, and it's a common type of problem you might encounter in your calculus studies. So, grab your thinking caps, and let's get started!

This problem beautifully illustrates the power of calculus in solving geometric optimization problems. We are tasked with finding the points on the curve defined by the equation 5x² - 6xy + 5y² = 4 that are closest to the origin. This involves several key concepts from calculus and analytic geometry, including distance formulas, optimization techniques, and the properties of conic sections. To successfully tackle this, we will need to formulate an objective function that represents the distance from a point on the curve to the origin, and then use calculus to minimize this function subject to the constraint given by the curve's equation. The method of Lagrange multipliers is particularly well-suited for this type of problem, as it allows us to handle constraints elegantly. By the end of this exploration, you will have a solid understanding of how to apply calculus to find minimum distances and how these techniques can be generalized to other optimization scenarios involving curves and surfaces. The blend of algebraic manipulation and calculus principles makes this a rich and rewarding problem to solve.

The challenge before us is not just about crunching numbers; it’s about strategically applying calculus concepts to a geometric problem. We need to think about how to translate the geometric requirement—finding the shortest distance—into a mathematical form that we can optimize. This involves defining the distance function, understanding the constraint imposed by the equation of the curve, and then choosing an appropriate method to handle this constrained optimization. The symmetry of the curve’s equation suggests that there might be multiple points that satisfy the condition of being closest to the origin, and our solution should be comprehensive enough to capture all such points. We will also need to verify that the points we find indeed correspond to minima and not maxima or saddle points, which is a crucial step in any optimization problem. By working through this problem, we’re not just finding an answer; we’re honing our problem-solving skills and deepening our understanding of the interplay between algebra, geometry, and calculus. The techniques we develop here will be valuable in a wide range of applications, from engineering design to economic modeling.

1. Setting Up the Problem

Okay, so the first thing we need to do is translate this geometric problem into math. What does it mean to find the points nearest the origin? Well, it means we want to minimize the distance between a point (x, y) on the curve and the origin (0, 0). Remember the distance formula? The distance d between (x, y) and (0, 0) is given by:

d = √(x² + y²)

Now, to make things a bit easier, instead of minimizing the distance d, we can minimize the square of the distance, d², which is:

d² = x² + y²

Minimizing d² will give us the same (x, y) points as minimizing d, but it gets rid of that pesky square root, making our calculations simpler. This is a common trick in optimization problems, guys!

Our goal is to minimize the square of the distance, d² = x² + y², from a point (x, y) on the curve to the origin. However, we can't just minimize x² + y² without any restrictions. The point (x, y) must lie on the curve defined by the equation 5x² - 6xy + 5y² = 4. This equation is our constraint. It tells us that x and y are not independent; they must satisfy this relationship. Therefore, we have a constrained optimization problem: minimize x² + y² subject to the constraint 5x² - 6xy + 5y² = 4. This is a classic setup for using the method of Lagrange multipliers, which is specifically designed to handle these types of constrained optimization problems. The Lagrange multiplier method allows us to find the stationary points of a function subject to one or more constraints by introducing a new variable (the Lagrange multiplier) and forming a new function (the Lagrangian). By setting the partial derivatives of the Lagrangian to zero, we obtain a system of equations that we can solve to find the candidate points for the minimum distance. This approach not only provides a systematic way to solve the problem but also gives us insights into the geometric relationship between the curve and the origin.

To set up the problem properly, we first need to clearly define our objective function and our constraint. Our objective function, which we want to minimize, is the square of the distance from the origin, f(x, y) = x² + y². This function represents the quantity we are trying to make as small as possible. Our constraint is the equation of the curve, g(x, y) = 5x² - 6xy + 5y² = 4. This equation restricts the possible values of x and y to only those that lie on the curve. The method of Lagrange multipliers tells us that at the points where the objective function is minimized (or maximized) subject to the constraint, the gradient of the objective function is parallel to the gradient of the constraint function. Mathematically, this means that there exists a scalar λ (the Lagrange multiplier) such that ∇f = λ∇g. This condition leads to a system of equations that we can solve to find the critical points. Setting up the problem in this way allows us to systematically find the points on the curve that are closest to the origin by transforming a geometric problem into an algebraic one. The elegance of this method lies in its ability to handle constraints seamlessly, providing a powerful tool for optimization in various fields.

2. Using Lagrange Multipliers

Here's where the magic happens! We'll use the method of Lagrange multipliers to solve this optimization problem. This method is perfect for finding the maximum or minimum of a function subject to a constraint. Basically, we introduce a new variable (the Lagrange multiplier), usually denoted by λ (lambda), and form a new function called the Lagrangian. The Lagrangian, L, is defined as:

L(x, y, λ) = f(x, y) - λ[g(x, y) - c]

Where:

f(x, y) is the function we want to minimize (in our case, x² + y²).
g(x, y) is the constraint function (in our case, 5x² - 6xy + 5y²).
c is the constant value of the constraint (in our case, 4).

So, for our problem, the Lagrangian becomes:

L(x, y, λ) = x² + y² - λ(5x² - 6xy + 5y² - 4)

Now, the trick is to find the critical points of L. We do this by taking the partial derivatives of L with respect to x, y, and λ, and setting them equal to zero:

∂L/∂x = 0
∂L/∂y = 0
∂L/∂λ = 0

Lagrange multipliers provide a powerful method for solving constrained optimization problems, where we seek to maximize or minimize a function subject to one or more constraints. The core idea behind this method is to transform the constrained optimization problem into an unconstrained one by introducing a new variable, the Lagrange multiplier λ, for each constraint. This approach is particularly useful when the constraint is given by an equation, as in our case with the curve 5x² - 6xy + 5y² = 4. The Lagrangian function, L(x, y, λ) = f(x, y) - λ[g(x, y) - c], combines the objective function f(x, y), which we want to minimize, and the constraint function g(x, y), ensuring that any solution we find satisfies the constraint. The constant c represents the value of the constraint, which is 4 in our problem. By taking the partial derivatives of the Lagrangian with respect to x, y, and λ, and setting them to zero, we obtain a system of equations. These equations represent the necessary conditions for a critical point of the Lagrangian, which corresponds to a point where the objective function is minimized or maximized subject to the constraint. The solutions to this system of equations give us the candidate points (x, y) and the corresponding Lagrange multiplier λ. It's important to note that not all solutions will necessarily be minima; we might also find maxima or saddle points. Therefore, further analysis is often needed to determine the nature of the critical points. The method of Lagrange multipliers is a cornerstone of optimization theory and has applications in various fields, including economics, engineering, and physics.

The effectiveness of the Lagrange multiplier method lies in its ability to incorporate the constraint directly into the optimization process. By forming the Lagrangian function, we are essentially creating a new function that implicitly includes the constraint. The partial derivatives of the Lagrangian with respect to x and y give us the conditions under which the gradient of the objective function is parallel to the gradient of the constraint function. This geometric interpretation is crucial for understanding why the method works. The condition ∂L/∂λ = 0 simply recovers the original constraint equation, ensuring that the solutions we find lie on the curve. Solving the system of equations resulting from setting these partial derivatives to zero can be challenging, often requiring algebraic manipulation and substitution. However, the systematic approach provided by the Lagrange multiplier method allows us to tackle complex optimization problems in a structured way. After finding the critical points, it is essential to verify that they correspond to minima, maxima, or saddle points by analyzing the second derivatives or using other methods. This verification step ensures that we have indeed found the points that are closest to the origin. The Lagrange multiplier λ itself can provide valuable information about the sensitivity of the objective function to changes in the constraint, making this method not only a problem-solving tool but also a source of deeper insights into the optimization problem.

3. Calculating the Partial Derivatives

Let's get our hands dirty with some calculus! We need to calculate those partial derivatives I mentioned. Remember, a partial derivative is just the derivative of a function with respect to one variable, treating all other variables as constants. So, let's do it:

∂L/∂x = ∂/∂x [x² + y² - λ(5x² - 6xy + 5y² - 4)] = 2x - λ(10x - 6y)
∂L/∂y = ∂/∂y [x² + y² - λ(5x² - 6xy + 5y² - 4)] = 2y - λ(-6x + 10y)
∂L/∂λ = ∂/∂λ [x² + y² - λ(5x² - 6xy + 5y² - 4)] = -(5x² - 6xy + 5y² - 4)

Now, we set each of these partial derivatives equal to zero:

2x - λ(10x - 6y) = 0
2y - λ(-6x + 10y) = 0
5x² - 6xy + 5y² - 4 = 0

This gives us a system of three equations with three unknowns (x, y, and λ). Our next step is to solve this system. Buckle up; it might get a bit algebraic!

Calculating the partial derivatives of the Lagrangian is a crucial step in the method of Lagrange multipliers. These derivatives provide us with the necessary conditions for finding the critical points of the Lagrangian, which correspond to the potential minima and maxima of our objective function subject to the constraint. The partial derivative with respect to x, ∂L/∂x, tells us how the Lagrangian changes as x varies, holding y and λ constant. Similarly, ∂L/∂y tells us how the Lagrangian changes as y varies, holding x and λ constant. The partial derivative with respect to λ, ∂L/∂λ, recovers our original constraint equation, ensuring that the solutions we find lie on the curve defined by 5x² - 6xy + 5y² = 4. Setting these partial derivatives to zero gives us a system of three equations with three unknowns (x, y, and λ), which we must solve to find the critical points. This system of equations represents the conditions where the gradient of the objective function is parallel to the gradient of the constraint function, a key concept in constrained optimization. The algebraic manipulations required to solve this system can be intricate, often involving substitution, elimination, and the use of symmetry to simplify the equations. The solutions we obtain will be the candidate points for the minimum distance from the origin to the curve, but we must remember that further analysis is needed to confirm that these points indeed correspond to minima and not maxima or saddle points. The accuracy of these derivative calculations is paramount, as any error here will propagate through the rest of the solution process. Therefore, careful attention to detail is essential in this step.

After obtaining the partial derivatives and setting them equal to zero, we are faced with the task of solving a system of three equations. This is where algebraic skill and strategic thinking become crucial. The equations we have are:

2x - λ(10x - 6y) = 0
2y - λ(-6x + 10y) = 0
5x² - 6xy + 5y² - 4 = 0

Solving this system requires a combination of techniques, such as substitution, elimination, and sometimes clever manipulation of the equations. For instance, we can rewrite equations 1 and 2 as:

2x = λ(10x - 6y)
2y = λ(-6x + 10y)

and then divide one equation by the other to eliminate λ, resulting in a relationship between x and y. This relationship can then be substituted into equation 3 to obtain an equation in a single variable. Alternatively, we can look for symmetry in the equations to simplify the system. The symmetry in the coefficients of x² and y² in the constraint equation suggests that there might be solutions where x = y or x = -y. Exploring these cases can often lead to a simpler set of equations to solve. The process of solving this system is not just about finding numerical values; it’s about developing a systematic approach to problem-solving, recognizing patterns, and making strategic decisions about which technique to apply. The algebraic skills honed in this step are invaluable in many areas of mathematics and its applications.

4. Solving the System of Equations

Alright, let's tackle this system of equations. This is often the trickiest part of Lagrange multiplier problems, but don't worry, we'll get through it together! First, let's rewrite equations (1) and (2) to make them a bit easier to work with:

2x = λ(10x - 6y)
2y = λ(-6x + 10y)

Now, a clever move here is to divide equation (1) by equation (2). This will eliminate λ, which is great! We get:

(2x) / (2y) = [λ(10x - 6y)] / [λ(-6x + 10y)]

Simplifying this, we get:

x/y = (10x - 6y) / (-6x + 10y)

Cross-multiplying gives us:

x(-6x + 10y) = y(10x - 6y)

Expanding, we get:

-6x² + 10xy = 10xy - 6y²

Notice that the 10xy terms cancel out! This simplifies our equation to:

-6x² = -6y²

Dividing both sides by -6, we get:

x² = y²

Taking the square root of both sides, we find:

x = ±y

This is a significant result! It tells us that our solutions will lie on the lines y = x and y = -x. Now, we need to use this information along with our constraint equation (3) to find the specific points.

The step-by-step solution of the system of equations is a critical part of the Lagrange multiplier method. The algebraic manipulations required here often demand creativity and a keen eye for simplification. By rewriting the equations and dividing them strategically, we managed to eliminate the Lagrange multiplier λ, which significantly simplified the problem. This step highlights the importance of looking for opportunities to reduce the number of variables in a system of equations. The resulting equation, x² = y², is a pivotal finding because it drastically narrows down the possible solutions. It tells us that the x-coordinate and the y-coordinate of the points closest to the origin are either equal or opposites of each other. This corresponds to two lines, y = x and y = -x, in the xy-plane. These lines represent the potential locations of the points on the curve that are closest to the origin. The next step involves using this information in conjunction with the constraint equation to find the specific points where these lines intersect the curve. The process of solving the system of equations is not just a mechanical procedure; it requires careful attention to detail and a willingness to explore different algebraic techniques. The ability to simplify complex equations is a valuable skill in many areas of mathematics and its applications, and this problem provides an excellent opportunity to hone this skill. The systematic approach we have taken, breaking the problem down into smaller steps, is key to successfully navigating the algebraic complexities of this type of problem.

The finding that x = ±y is a significant breakthrough in our problem-solving journey. This result geometrically represents two lines, y = x and y = -x, which pass through the origin and divide the coordinate plane into four equal parts. These lines are the potential locations of the points on the curve 5x² - 6xy + 5y² = 4 that are closest to the origin. The symmetry of these lines about the origin reflects the symmetry of the curve's equation, which is a crucial observation. This symmetry suggests that if a point (x, y) on the curve is closest to the origin, then the point (-x, -y) might also be a solution. Now, to find the specific points, we need to substitute these relationships into the constraint equation 5x² - 6xy + 5y² = 4. This will give us two separate cases to consider:

Case 1: y = x
Case 2: y = -x

Each of these cases will lead to a quadratic equation in x, which we can solve to find the x-coordinates of the points of intersection. Once we have the x-coordinates, we can use the relationships y = x and y = -x to find the corresponding y-coordinates. This systematic approach ensures that we consider all possible solutions and that we use the information we have gained so far in the most efficient way. The ability to break a problem down into smaller, more manageable cases is a key strategy in problem-solving, and this example beautifully illustrates this technique.

5. Substituting Back into the Constraint Equation

Now that we know x = ±y, let's plug these cases back into our constraint equation, 5x² - 6xy + 5y² = 4, to find the specific points.

Case 1: y = x

Substituting y = x into the constraint equation, we get:

5x² - 6x(x) + 5(x)² = 4

Simplifying, we have:

5x² - 6x² + 5x² = 4

4x² = 4

Dividing both sides by 4, we get:

x² = 1

Taking the square root, we find:

x = ±1

Since y = x, the corresponding y values are also ±1. So, we have two points: (1, 1) and (-1, -1).

Case 2: y = -x

Substituting y = -x into the constraint equation, we get:

5x² - 6x(-x) + 5(-x)² = 4

Simplifying, we have:

5x² + 6x² + 5x² = 4

16x² = 4

Dividing both sides by 16, we get:

x² = 1/4

Taking the square root, we find:

x = ±1/2

Since y = -x, the corresponding y values are ∓1/2. So, we have two more points: (1/2, -1/2) and (-1/2, 1/2).

Now we have four candidate points: (1, 1), (-1, -1), (1/2, -1/2), and (-1/2, 1/2). The final step is to determine which of these points are actually closest to the origin.

Substituting the relationships y = x and y = -x back into the constraint equation is a crucial step in finding the specific points that satisfy both the equation of the curve and the condition of being closest to the origin. This process involves careful algebraic manipulation and simplification. In Case 1, where y = x, the substitution leads to the equation 4x² = 4, which simplifies to x² = 1. This gives us two solutions for x: x = 1 and x = -1. Since y = x, the corresponding y-values are also 1 and -1, respectively. This yields the points (1, 1) and (-1, -1). In Case 2, where y = -x, the substitution leads to the equation 16x² = 4, which simplifies to x² = 1/4. This gives us two solutions for x: x = 1/2 and x = -1/2. Since y = -x, the corresponding y-values are -1/2 and 1/2, respectively. This yields the points (1/2, -1/2) and (-1/2, 1/2). These four points are the candidates for the points on the curve that are closest to the origin. However, we still need to verify which of these points actually minimize the distance to the origin. This can be done by calculating the distance from each of these points to the origin and comparing the results. The algebraic precision in this substitution and simplification process is essential to ensure that we obtain the correct candidate points. The systematic approach of considering each case separately helps to avoid errors and ensures that we do not miss any potential solutions. The next step involves evaluating the objective function (the square of the distance) at these candidate points to determine which ones are closest to the origin.

After substituting the relationships x = ±y into the constraint equation, we have identified four candidate points: (1, 1), (-1, -1), (1/2, -1/2), and (-1/2, 1/2). These points are the potential locations on the curve 5x² - 6xy + 5y² = 4 that are closest to the origin. To determine which of these points are indeed the closest, we need to evaluate the objective function, which is the square of the distance from the origin, d² = x² + y², at each of these points. This will allow us to compare the distances and identify the points that minimize the distance to the origin. The evaluation process is straightforward: we simply substitute the x and y coordinates of each point into the equation d² = x² + y² and calculate the result. The point(s) with the smallest value of d² will be the closest to the origin. This evaluation step is crucial because not all critical points found using Lagrange multipliers correspond to minima; some may be maxima or saddle points. Therefore, we must explicitly compare the values of the objective function at the candidate points to ensure that we identify the points that truly minimize the distance. The systematic evaluation of these candidate points is a key part of the optimization process, and it demonstrates the importance of not only finding critical points but also verifying their nature.

6. Determining the Minimum Distance

Okay, we have our candidate points. Now, we need to figure out which ones are actually closest to the origin. Remember, we're minimizing d² = x² + y². So, let's calculate d² for each point:

For (1, 1): d² = 1² + 1² = 2
For (-1, -1): d² = (-1)² + (-1)² = 2
For (1/2, -1/2): d² = (1/2)² + (-1/2)² = 1/4 + 1/4 = 1/2
For (-1/2, 1/2): d² = (-1/2)² + (1/2)² = 1/4 + 1/4 = 1/2

Comparing these values, we see that the minimum value of d² is 1/2, which occurs at the points (1/2, -1/2) and (-1/2, 1/2). Therefore, these are the points on the curve that are nearest the origin!

So, guys, we've successfully found the points on the curve 5x² - 6xy + 5y² = 4 that are closest to the origin. The points are (1/2, -1/2) and (-1/2, 1/2). This problem was a great exercise in using Lagrange multipliers and combining calculus with geometric concepts. I hope you found this explanation helpful. Keep practicing, and you'll become a calculus pro in no time!

Determining the minimum distance involves evaluating the objective function, which in our case is the square of the distance from the origin (d² = x² + y²), at each of the candidate points we found. This is a crucial step in any optimization problem, as it allows us to compare the values of the objective function and identify the points that yield the minimum (or maximum) value. For the point (1, 1), we have d² = 1² + 1² = 2. For the point (-1, -1), we have d² = (-1)² + (-1)² = 2. For the point (1/2, -1/2), we have d² = (1/2)² + (-1/2)² = 1/4 + 1/4 = 1/2. For the point (-1/2, 1/2), we have d² = (-1/2)² + (1/2)² = 1/4 + 1/4 = 1/2. By comparing these values, we can see that the minimum value of d² is 1/2, which occurs at the points (1/2, -1/2) and (-1/2, 1/2). This means that these points are the closest to the origin among the candidate points we identified. The corresponding minimum distance d is the square root of 1/2, which is √(1/2) or 1/√2. This evaluation step not only confirms the minimum distance but also highlights the symmetry of the problem, as we found two points that are equidistant from the origin and lie on the curve. The careful calculation of the distances and the comparison of their values are essential to ensure that we have correctly identified the points that solve our optimization problem. The clear and systematic approach we have taken, from setting up the problem to evaluating the candidate points, demonstrates the power of calculus in solving geometric optimization problems.

Conclusion

In conclusion, by systematically applying the method of Lagrange multipliers, we have successfully found the points on the curve 5x² - 6xy + 5y² = 4 that are nearest to the origin. These points are (1/2, -1/2) and (-1/2, 1/2). This problem demonstrates the power of calculus in solving geometric optimization problems and highlights the importance of techniques such as Lagrange multipliers. The process involved setting up the problem, forming the Lagrangian, calculating partial derivatives, solving a system of equations, and evaluating candidate points to determine the minimum distance. Each of these steps required careful attention to detail and a solid understanding of the underlying mathematical principles. The solution not only provides the answer to the specific problem but also offers insights into the geometric properties of the curve and the nature of constrained optimization. The symmetry observed in the solution reflects the symmetry of the curve's equation, which is a common theme in mathematical problems. By mastering these techniques, you will be well-equipped to tackle a wide range of optimization problems in various fields, from engineering to economics. The journey from the initial problem statement to the final solution underscores the elegance and effectiveness of calculus as a problem-solving tool.