Question

How do you find the vertex, focus and directrix of `y^2=-8x`?

Answer 1

The vertex is at the point `(0,0)`
The focus is at the point `(-2, 0)`
The equation of the directrix line is `x = 2`

Explanation:

Let's begin by writing this in the form, `x = ay² + by + c`:

`x =(-1/8)y²`

Please recognize that this is a parabola that is rotated counterclockwise 90° so that it opens to the left. We know that it is rotated, because it uses "y" as the dependent variable. We know that it opens to the left, because the value of "a" is negative; if the value of "a" were positive, then we would conclude that it opens to the right.

The y coordinate of the vertex is found using the equation:

`y = -b/(2a)`

`y = -0/(2(-1/8))`

`y = 0`

The corresponding x coordinate is found by evaluating the given equation at the y coordinate:

`x = (-1/8)0²`

`x = 0`

A parabola of this type has its focus to the left of the vertex and the focus has the same y coordinate as the vertex. The signed distance from the vertex to the focus is found by using the equation:

`f = 1/(4a)`

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

`f = -2`

The y coordinate of the focus is found, using the equation:

`x = x_v + f`

where `x_v` is the x coordinate of the vertex:

`x = 0 + -2`

`x = -2`

The vertex is at the point `(-2, 0)`

The directrix is a vertical line to the right of the vertex. The equation for the directrix line is found using the equation:

`x = x_v - f`

`x = 0 - -2`

`x = 2`