Question

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

Answer 1

Explanation:

C) y = –1∕4 (x – 3)^2 + 2

Answer 2

C

Explanation:

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

Using the distance formula

`√((x-3)^2+(y-1)^2)` = | 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 )

-
`(1)/(4)` (x - 3)² + 2 = y, that is

y = -
`(1)/(4)` (x - 3)² + 2 → C