# How do you identify the vertex, focus, directrix and the length of the latus rectum and graph 4(x-2)=(y+3)^2?

Oct 30, 2016

#### Explanation:

Rewrite the equation in the vertex form

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

where $\left(h , k\right)$ is the vertex and the signed distance of the vertex to the focus is $f = \frac{1}{4 a}$

Divide by both sides by 4:

$x - 2 = \frac{1}{4} {\left(x + 3\right)}^{2}$

Write the + as a - -

$x - 2 = \frac{1}{4} {\left(x - - 3\right)}^{2}$

$x = \frac{1}{4} {\left(y - - 3\right)}^{2} + 2$

Now, that the equation is in standard form, please observe that the vertex is at $\left(2 , - 3\right)$

The distance, in the x direction, from the vertex from to the focus is:

$f = \frac{1}{4 \left(\frac{1}{4}\right)} = 1$

Therefore, we add 1 to the x coordinate of the vertex and we see that the focus is $\left(3 , - 3\right)$

The directrix is the vertical line the same distance in the opposite direction from the vertex so we subtract 1 from the x coordinate of the vertex; making its equation:

$x = 1$

Modifying the equation, $f = \frac{1}{4 a}$, to be $a = \frac{1}{4 f}$, the length of the latus rectum is the denominator, $4 f$:

$4 f = 4 \left(1\right) = 4$