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

Answer 1

Lincoln Anderson

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#

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Answer 2

Benjamin Achtenberg

To find the focus, vertex, and directrix of the parabola represented by the equation x = 3(y – 1)^2 + 2, you need to rewrite the equation in standard form (y - k)^2 = 4p(x - h), where (h, k) is the vertex and p is the distance from the vertex to the focus and from the vertex to the directrix. Then, you can easily identify the focus, vertex, and directrix.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.