# 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 = √(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{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} + y^{2} - 2y + 1 = y^{2} + 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].

## 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].

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