find the max and min distance from the origin to the ellipse x^2+xy+y^2=3<hint use x^2+y^2 as your objective function
https://math.stackexchange.com/questions/214078/find-a-point-on-ellipse-closest-to-origin
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To find the maximum and minimum distance from the origin to the ellipse x^2 + xy + y^2 = 3, we can use the concept of optimization along with the given objective function x^2 + y^2. Let's go step by step:
1. Start by rewriting the equation of the ellipse in terms of the objective function:
x^2 + xy + y^2 = 3
Rearranging this equation, we get:
x^2 + y^2 = 3 - xy
2. Now, our objective function is x^2 + y^2, which represents the distance squared from the origin. We square the distance to avoid dealing with square roots during optimization.
3. The next step is to maximize and minimize the objective function x^2 + y^2, subject to the constraint of the ellipse equation. This can be done using Lagrange multipliers.
4. Formulate the Lagrangian function L(x, y, λ) as follows:
L(x, y, λ) = x^2 + y^2 + λ(3 - xy)
5. Next, find the partial derivatives of the Lagrangian function with respect to x, y, and λ:
∂L/∂x = 2x - λy
∂L/∂y = 2y - λx
∂L/∂λ = 3 - xy
6. Set the partial derivatives equal to zero to find the critical points:
∂L/∂x = 2x - λy = 0
∂L/∂y = 2y - λx = 0
∂L/∂λ = 3 - xy = 0
7. Solve the system of equations formed by setting the partial derivatives equal to zero. This will give us the critical points.
8. Once you find the critical points, substitute them back into the objective function x^2 + y^2 to get the values of the objective function at those points.
9. The maximum and minimum values of the objective function obtained from the critical points will correspond to the maximum and minimum distances from the origin to the ellipse.
10. Calculate the square root of the maximum and minimum values obtained in step 9 to obtain the actual maximum and minimum distances from the origin to the ellipse.