Question

Find the points on the ellipse 4x2 + y2 = 4 that are farthest away from the point (1, 0).

Answer 1

4x² + y² = 4 ──► y² = 4 - 4x² ──► y = ±√(4 - 4x²)

Now use thedistance formula:
D = √((x-1)² + (y-0)²)
= √((x-1)² + y²)
= √(x² - 2x + 1 + 4 - 4x²)
= √(-3x² - 2x + 5)

Now take the derivative of that and set it equal to 0:
dD/dx = (½)(-3x² - 2x + 5)^(-½)(-6x - 2)
= (-6x - 2) / (2·√(-3x² - 2x + 5)) = 0

Multiply both sides by the denominator:
-6x - 2 = 0
-6x = 2
x = -⅓

Plug this back into the ellipse equation and solve for y:
y = ±√(4 - 4( -⅓)²) = ±√(4 - 4/9) ≈ ±1.89

So the solution set is:
(x,y) = {(-0.33, -1.89), (-0.33, 1.89)}