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

Dec 2, 2016

#### Explanation:

Given: $4 x - {y}^{2} - 2 y - 33 = 0$

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

$4 x = {y}^{2} + 2 y + 33$

Divide both side by 4:

$x = \frac{1}{4} {y}^{2} + \frac{1}{2} y + \frac{33}{4} \text{ [1]}$

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:

$x = a {\left(y - k\right)}^{2} + h$

where "a" is the coefficient of the ${y}^{2}$ term and $\left(h , k\right)$ is the vertex.

The focus of this type is located at $\left(h + \frac{1}{4 a} , k\right)$

The equation of the directrix is $x = h - \frac{1}{4 a}$

Lets put equation [1] in vertex from. Add zero to equation [1] in the form of $\frac{1}{4} {k}^{2} - \frac{1}{4} {k}^{2}$:

$x = \frac{1}{4} {y}^{2} + \frac{1}{2} y + \frac{1}{4} {k}^{2} - \frac{1}{4} {k}^{2} + \frac{33}{4}$

Factor $\frac{1}{4}$ from the first 3 terms:

$x = \frac{1}{4} \left({y}^{2} + 2 y + {k}^{2}\right) - \frac{1}{4} {k}^{2} + \frac{33}{4} \text{ [2]}$

Please observe that the right side of the pattern ${\left(y - k\right)}^{2} = {y}^{2} - 2 k + {k}^{2}$ 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]:

$- 2 k = 2 y$

$k = - 1$

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

$x = \frac{1}{4} {\left(y - k\right)}^{2} - \frac{1}{4} {k}^{2} + \frac{33}{4}$

Substitute -1 for every k:

$x = \frac{1}{4} {\left(y - - 1\right)}^{2} - \frac{1}{4} {\left(- 1\right)}^{2} + \frac{33}{4}$

Simplify the constant term:

$x = \frac{1}{4} {\left(y - - 1\right)}^{2} + 8$

The vertex is at $\left(8 , - 1\right)$

The focus is at

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

This simplifies to:

(9, - 1)#

The equation of the directrix is

$x = 8 - \frac{1}{4 \left(\frac{1}{4}\right)}$

$x = 7$