Question

What is the focus of the parabola given by the equation x = 3(y – 1)^2 + 2?

Answer 1

i believe itll be related to x = 2y

Answer 2

maybe x = 2(y-1) +2 to be centered right tho

Answer 3

i forget what the geometric equations for these are ...

Answer 4

3(y – 1)^2 + 2 = 2(y – 1) + 2

3(y – 1)^2 = 2(y – 1)

3(y – 1) = 2

y = 2/3 + 1 = 5/3, but id need to wolf it to be sure

Answer 5

so it has a squared "y" component, thus the "focus form" of it will be \(\bf (y-k)^2=4p(x-h)\)

(h, k) = vertex point
p = distance from the vertex to the focus
p < 0, opens to the left
p > 0, opens to the right

Answer 6

if im right ... and i seldom am: (2,5/3) maybe?

Answer 7

yeah, im wrong .... its just turned all backwards is all

Answer 8

well, came out as \(\bf x = 3(y-1)^2+2 \implies \cfrac{1}{3}(x-2) = (y-1)^2\\
4p = \cfrac{1}{3} \implies p = \cfrac{1}{12}\)

Answer 9

of the answer choices there is (2, 1 1/12) and (2 1/12, 1) those are the only that have 1/12

Answer 10

so the vertex is at (2, 1)
so the focus will be at \(\bf \left(2+ \cfrac{1}{12}, 1\right)\)