Find the points on the ellipse 81x^2+y^2=81 that are farthest away from the point (1,0). List in the form of (a,b)

Answer 1

Answer 2

√The distance from some point (x, y) that is on the ellipse can be specified as:

d = √[(x - x₁)² + (y - y₁)²] where (x₁, y₁) is an arbitrary point, in this case (1, 0). So, inserting the point we get:

d = √[(x - 1)² + (y - 0)²] or

d = √[(x - 1)² + y²]

But since 81x² + y² = 81 then y² = 81 - 81x² so, plugging in "81 - 81x²" for "y²" into the distance formula we get:

d(x) = √[(x - 1)² + 81 - 81x²] or

d(x) = √[x² - 2x + 1 + 81 - 81x²] or

d(x) = √[82 - 2x - 80x²]

To find the maximum distance, take the derivative and set it equal to zero.

d'(x) = 1/2(82 - 2x - 80x²)^-1/2(-2 - 160x) or

d'(x) = (-1 - 80x)/(82 - 2x - 80x²)^1/2

This is equal to zero when the numerator equals zero so:

-1 - 80x = 0

-80x = 1

x = -1/80

plugging this into y² = 81 - 81x² we get:

y² = 81 - 81(-1/80)²

y² = 81 - 81/6400

y = ± √518319/6400 = √518319/80 ≈8.9993

So the points farthest away from (1, 0) on the ellipse 81x² + y² = 81 are (-1/80, 8.9993) and (-1/80, -8.9993)