Question

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

Answer 1

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`