Question

How do you find the focus, directrix and sketch `x=2y^2-4`?

Answer 1

The focus will be shifted horizontally by the distance `f` from the vertex.
The directrix will be vertical line and shifted a distance `-f` from the vertex.

Explanation:

Because the coefficient of the y term is zero, we know that the y coordinate of the vertex is:

`k = 0`

The x coordinate, h, is found by evaluating the equation at `y = k = 0`:

`h = 2(0)^2-4`

`h = -4`

We know that the focal distance, f, is the signed horizontal shift from the vertex to the focus and it can be found using the coefficient of the `y^2` term:

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

In our case `a = 2`

`f = 1/8`

The x coordinate of the focus is:

`x_f = h + f`

`x_f = -4+1/8`

`x_f = 31/8`

We know that the y coordinate of the focus is the same as the vertex, therefore the focus is `(31/8,0)`

The equation of the directrix is:

`x = h-f`

`x = -4-1/8`

`x = -33/8`

Here is a graph:

graph{x=2y^2-4 [-10, 10, -5, 5]}