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 = √(x2 - x1)2 + (y2 - y1)2
Distance A and F is dAF = √(x - 3)2 + (y - (-4))2 = √(x - 3)2 + (y + 4)2
Distance between A and D is dAD = √(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 + y2 + 8y + 16 = y2 - 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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