Question

Derive the equation of a parabola with a focus at (6,−2)
and a directrix at y=0
.(1 point)

y=−14x2+2x−6


y=−14x2+3x−10


y=14x2+x+4


y=14x2+3x+8

Answer 1

Focus (6, −2) and directrix y = 0 => vertex midway: (6, −1). Focal parameter p = −1 (focus is 1 below vertex). Standard form:
(x − 6)^2 = 4p(y + 1) = −4(y + 1).

Solve for y:
(x − 6)^2 = −4(y + 1) ⇒ y = −1/4 x^2 + 3x − 10.

This matches option 2 (interpreting −14x^2 as −1/4 x^2).