Reviewed by Yaren Fadiloglulari
Find the points on the ellipse 4x^2 + y^2 = 4 that are farthest away from the point (-1, 0)
Answer: The points on the ellipse 4x^2 + y^2 = 4 that are farthest away from (-1, 0) are (1, 0) and (-1, 2)
This problem asks us to find the points located on the ellipse defined by the equation 4x^2 + y^2 = 4 that are the farthest away from the specific external point (-1, 0). To solve this, we need to maximize the distance from (-1, 0) to any point (x, y) on the ellipse, using the tools of calculus (specifically optimization coupled with constraint equations) or geometric reasoning.
Methods
Math Tutor Explanation Using the Distance Maximization Method
We solve this problem by maximizing the distance from (-1, 0) to points (x, y) on the ellipse, using calculus with constraints (the method of Lagrange multipliers).
Step 1: Step 1: Write the distance squared function D^2 = (x + 1)^2 + y^2, and the constraint 4x^2 + y^2 = 4
Step 2: Step 2: Set up the Lagrange multipliers equations: ∇D^2 = λ∇(ellipse constraint)
Math Tutor Explanation Using Substitution and Quadratic Maximization
By substituting y^2 = 4 - 4x^2 into the distance squared, we convert the problem to finding the maximum of a function in one variable.
Step 1: Step 1: Substitute y^2 = 4 - 4x^2 into D^2 = (x + 1)^2 + y^2, giving D^2(x) = (x + 1)^2 + 4 - 4x^2
Step 2: Step 2: Simplify D^2(x), yielding D^2(x) = -3x^2 + 2x + 5, and find where this quadratic attains its maximum within the allowable x-values
Step 1:
Step 2:
Math Tutor suggests: Dive Deeper into Ellipses and Distance Problems
Explore more problems related to ellipses, conic sections, and maximizing distances. Try these related questions to strengthen your skills.
FAQ on Ellipse Distance Optimization
How do I use Lagrange multipliers for this type of problem?
Set up the function to optimize along with the constraint, and solve the resulting system of equations involving partial derivatives.
Can you have more than one farthest point on an ellipse from a fixed point?
Yes, symmetrical or geometrical considerations can lead to multiple points with equal maximal distances.
Why do you maximize the square of the distance instead of the distance itself?
Maximizing the square of the distance avoids dealing with square roots and leads to the same location results.
Do optimization techniques work for other conic sections?
Yes, similar methods can be applied to circles, hyperbolas, and parabolas.
What if the point is inside the ellipse?
The procedures are similar, but the range of distances and possible extremal points may change.
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