The minimum distance from the parabola y = x² to the point (0, 9) is 9 units. This distance occurs at two points: (3, 9) and (-3, 9).
Here's how to find the minimum distance from the parabola y = x² to the point (0, 9) using Lagrange multipliers:
1. Set up the problem:
Define the distance function to minimize: D(x, y) = √((x - 0)² + (y - 9)²)
Define the constraint function for the parabola: g(x, y) = y - x²
2. Apply Lagrange multipliers:
Form the Lagrangian function: L(x, y, λ) = D(x, y) + λ * g(x, y)
Substitute the distance and constraint functions: L(x, y, λ) = √((x - 0)² + (y - 9)²) + λ (y - x²)
3. Find the critical points:
Take partial derivatives of L with respect to x, y, and λ, and set them equal to 0:
∂L/∂x = (x - 0) / √((x - 0)² + (y - 9)²) - 2λx = 0
∂L/∂y = (y - 9) / √((x - 0)² + (y - 9)²) + λ = 0
∂L/∂λ = y - x² = 0
4. Solve the system of equations:
From the third equation, y = x². Substitute this into the first and second equations:
x² / √(x² + (x² - 9)²) - 2λx = 0
x² / √(x² + (x² - 9)²) + λ = 0
Simplify and solve for x and λ:
x^4 = 27λx^2
λ = 9/x^2 (from the second equation)
Substitute the second equation into the first and solve for x: x^4 = 27 (9/x^2)x^2
This simplifies to x^6 = 243, which has two solutions: x = 3 and x = -3
5. Find the minimum distance:
Substitute the two solutions for x back into the distance function:
D(3, 9) = √((3 - 0)² + (9 - 9)²) = 9
D(-3, 9) = √((-3 - 0)² + (9 - 9)²) = 9
6. Conclusion:
The minimum distance from the parabola y = x² to the point (0, 9) is 9 units. This distance occurs at two points: (3, 9) and (-3, 9).