How do you find the vertex, focus and directrix of x = y^2 - 6y + 11?

1 Answer
Nov 26, 2016

Please see the explanation.

Explanation:

The vertex form of the equation of a parabola that opens left or right is:

x = a(y - k)^2 + h

where (h, k) is the vertex and a is the coefficient of the y^2 term.

To this end, add 0 to the given equation in the form k^2 - k^2:

x = y^2 -6y + k^2 - k^2 + 11

NOTE: In this case, a = 1. If it were something other than 1, we would substitute ak^2 - ak^2 and then remove the factor of "a" from the first 3 terms.

Set the middle term in the pattern, (y - k)^2 = y^2 - 2ky + k^2, equal to the corresponding term in the given equation:

-2ky = -6y

Solve for k:

k = 3

Substitute the left side of the pattern for the first 3 terms of the equation:

x = (y - k)^2 - k^2 + 11

NOTE: If "a" were something other than one, we would substitute into the parenthesis that "a" multiplies. Here is an example with a = 2: x = 2(y - k)^2 - 2k^2 + 11

Substitute 3 for k:

x = (y - 3)^2 - 3^2 + 11

Combine the constant terms:

x = (y - 3)^2 + 2

Obtain the vertex by observation: (2, 3)

The equation of the distance, f, from the vertex to the focus is:

f = 1/(4a)

Substitute 1 for a:

f = 1/4
The general form for the focus is:

(h + f, k)

The focus is at: (2.25, 3)

The directrix is a vertical line whose general equation is:

x = h - f

The directrix is:

x = 1.75