How do you find the focus, vertex, and directrix of $x = 3(y – 1)^2 + 2$ ?

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Answer 1 · 2016-10-03T15:35:19

The vertex is $(1, 2)$
The focus is $(1, 25/12)$
The equation of the directrix is: $x = 23/12$

The given equation is in the vertex form of a parabola that opens either to the left or right:

$x = 1/(4f)(y - k)² + h$

where (h, k) is the vertex and f is the distance from the vertex to the focus.

The vertex is $(1, 2)$

The focus has the same x coordinate as the vertex but the y coordinate is increased by distance f:

$3 = 1/(4f)$

$f = 1/12$

The focus is $(1, 25/12)$

The directrix is a vertical line with perpendicular distance -f from the vertex:

The equation of the directrix is: $x = 23/12$

How do you find the focus, vertex, and directrix of x = 3(y – 1)^2 + 2x = 3(y – 1)^2 + 2?