# Using a directrix of y = 2 and a focus of (3, -4), 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 = (3, -4)

Given, directrix = 2

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

First, find out the distance using distance formula,

d = √(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

Distance A and F is d_{AF} = √(x - 3)^{2} + (y - (-4))^{2} = √(x - 3)^{2} + (y + 4)^{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 - 3)^{2} + (y + 4)^{2} = √(y - 2)^{2}

Squaring both sides,

(√(x - 3)^{2} + (y + 4)^{2})^{2} = (√(y - 2)^{2})^{2}

(x - 3)^{2} + (y + 4)^{2} = (y - 2)^{2}

(x - 3)^{2 }+ y^{2} + 8y + 16 = y^{2} - 4y + 4.

(x - 3)^{2} + 8y + 4y + 16 - 4 = 0

(x - 3)^{2} + 12y + 12 = 0

12y = -(x - 3)^{2} - 12

y = -1/12 [(x - 3)^{2} + 1]

Therefore, the quadratic function is y = -1/12 [(x - 3)^{2} + 1].

## Using a directrix of y = 2 and a focus of (3, -4), what quadratic function is created?

**Summary:**

Using a directrix of y = 2 and a focus of (3, -4), the quadratic function created is f(x) = y = -1/12 [(x - 3)^{2} + 1].

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