Question

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

Answer 1

Please see the explanation.

Explanation:

Rewrite the equation in the vertex form

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

where `(h, k)` is the vertex and the signed distance of the vertex to the focus is `f= 1/(4a)`

Divide by both sides by 4:

`x - 2 = 1/4(x + 3)^2`

Write the + as a - -

`x - 2 = 1/4(x - -3)^2`

Add 2 to both sides:

`x = 1/4(y - -3)^2 + 2`

Now, that the equation is in standard form, please observe that the vertex is at `(2, -3)`

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

`f = 1/(4(1/4)) = 1`

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

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 = 1/(4a)`, to be `a = 1/(4f)`, the length of the latus rectum is the denominator, `4f`:

`4f = 4(1) = 4`