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16. Find the equation of a parabola with a focus at (3,1) and a directrix at y = 3.
A) y = 1∕4(x – 3)^2 + 3
B) y = –1∕4 (x – 3)^2 + 3
C) y = –1∕4 (x – 3)^2 + 2
D) y = 1∕4(x – 3)^2 + 2

Respuesta :

Answer:

C

Step-by-step explanation:

Any point (x, y) on the parabola is equidistant from the focus and the directrix.

Using the distance formula

[tex]\sqrt{(x-3)^2+(y-1)^2}[/tex] = | y - 3 | ← square both sides

(x - 3)² + (y - 1)² = (y - 3)² ← expand the y- factors

(x - 3)² + y² - 2y + 1 = y² - 6y + 9 ← subtract y² - 2y + 1 from both sides

(x - 3)² = - 4y + 8 ( subtract 8 from both sides )

(x - 3)² - 8 = - 4y ( divide both sides by - 4 )

- [tex]\frac{1}{4}[/tex] (x - 3)² + 2 = y, that is

y = - [tex]\frac{1}{4}[/tex] (x - 3)² + 2 → C

Answer:

Step-by-step explanation:

C) y = –1∕4 (x – 3)^2 + 2

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