Using a directrix of y = -3 and a focus of (2, 1), what quadratic function is created?

Solution:

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

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

Focus, F = (2,1)

Given, directrix = -3

D = (x, -3) 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 - 1)²

Distance between A and D is d_AD = √(x - x)² + (y - (-3))² = √(y + 3)²

Since these distances must be equal to each other,

√(x - 2)² + (y - 1)² = √(y + 3)²

Squaring both sides,

(√(x - 2)² + (y - 1)²)² = (√(y + 3)²)²

(x - 2)² + (y - 1)² = (y + 3)²

(x - 2)² + y² - 2y + 1 = y² + 6y + 9.

(x - 2)² + 1 - 9 - 2y - 6y = 0

(x - 2)² - 8 - 8y = 0

8y = (x - 2)² - 8

y = 1/8 [(x - 2)² - 1]

Therefore, the quadratic function is y = 1/8 [(x - 2)² - 1].

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Using a directrix of y = -3 and a focus of (2, 1), what quadratic function is created?

Summary:

Using a directrix of y = -3 and a focus of (2, 1), the quadratic function created is f(x) = 1/8 [(x - 2)² - 1].