Using a directrix of y = -2 and a focus of (2, 6), what quadratic function is created?
f(x) = -1/8(x - 2)² - 2
f(x) = 1/16(x - 2)² + 2
f(x) = 1/8(x - 2)² - 2
f(x) = -1/16(x + 2)² - 2 

Solution:

The definition of a parabola states that all points on the parabola always have the same distance to the focus and the directrix.

Given: Focus, F = (2, 6) and directrix = -2

Let A = (x, y) be a point on the parabola.

D = (x, -2) represent the closest point on the directrix

First, find out the distance using distance formula,

d = √(x₂ - x₁)² + (y₂ - y₁)²

Distance A and F is d_AF = √(x - 2)² + (y - 6)²

Distance between A and D is d_AD = √(x - x)² + (y - (-2))² = √(y + 2)²

Since these distances must be equal to each other, 

 √(x - 2)² + (y - 6)² = √(y + 2)²

Squaring both sides,

⇒ (√(x - 2)² + (y - 6)²)² = (√(y + 2)²)²

⇒ (x - 2)² + (y - 6)² = (y + 2)²

⇒ (x - 2)² + y² - 12y + 36 = y² + 4y + 4.

⇒ (x - 2)² + 36 - 4 - 12y - 4y = 0

⇒ (x - 2)² + 32 - 16y = 0

⇒ 16y = (x - 2)² + 32

⇒ y = 1/16 [(x - 2)² + 2]

Therefore, the quadratic function is y = 1/16 [(x - 2)² + 2].

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Using a directrix of y = -2 and a focus of (2, 6), what quadratic function is created?

Summary:

Using a directrix of y = -2 and a focus of (2, 6), the quadratic function created is f(x) = 1/16 [(x - 2)² + 2].