Question

How do you find the vertex, focus and directrix of `4x-y^2-2y-33=0`?

Answer 1

Please see the explanation.

Explanation:

Given: `4x - y^2 - 2y - 33 = 0`

Add #y^2 + 2y + 33 to both sides:

`4x = y^2 + 2y + 33`

Divide both side by 4:

`x = 1/4y^2 + 1/2y + 33/4" [1]"`

This type of parabola opens to left or right. Because the coefficient, a, of the `y^2` is greater than zero, we know that is opens to the right.

The vertex form of an equation of this type of parabola is:

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

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

The focus of this type is located at `(h + 1/(4a),k)`

The equation of the directrix is `x = h - 1/(4a)`

Lets put equation [1] in vertex from. Add zero to equation [1] in the form of `1/4k^2 - 1/4k^2`:

`x = 1/4y^2 + 1/2y + 1/4k^2 - 1/4k^2 + 33/4`

Factor `1/4` from the first 3 terms:

`x = 1/4(y^2 + 2y + k^2) - 1/4k^2 + 33/4" [2]"`

Please observe that the right side of the pattern `(y - k)^2 = y^2 -2k + k^2` matches what is inside the parenthesis in equation [2]. Set the middle term of the right side of the pattern equal to the corresponding term in equation [2]:

`-2k = 2y`

`k = -1`

Substitute the left side of the pattern into the ()s in equation [2]:

`x = 1/4(y - k)^2 - 1/4k^2 + 33/4`

Substitute -1 for every k:

`x = 1/4(y - -1)^2 - 1/4(-1)^2 + 33/4`

Simplify the constant term:

`x = 1/4(y - -1)^2 + 8`

The vertex is at `(8, -1)`

The focus is at

#(8 + 1/(4(1/4)), -1)

This simplifies to:

(9, - 1)#

The equation of the directrix is

`x = 8 - 1/(4(1/4))`

`x = 7`