Question

How do you find the vertex, focus and directrix of `2(x-3)^2=6y+72`?

Answer 1

The procedure for finding the vertex, focus, and directrix is contained in the explanation.

Explanation:

The vertex form for the equation of a parabola is:

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

where `(h, k)` is the vertex and `a` is the same as the first coefficient in the standard form, `y = ax^2 + bx + c`

Let's proceed with the steps to put the given equation in the vertex form:

`2(x - 3)^2 = 6y + 72`

`6y + 72 = 2(x - 3)^2`

`y + 16 = 1/3(x - 3)^2`

`y = 1/3(x - 3)^2 - 16`

The vertex is obtained by inspection; it is the point, `(3, - 16)`

The equation for the focal distance, f, is:

`f = 1/(4a)`

Substitute `1/3`for `a`:

`f = 1/(4(1/3))`

`f = 3/4`

The general form of the focal point is (h, k + f)

Substitute in our values:

`(3, -16 + 3/4)`

The focal point is:

`(3, -61/4)`

The directrix is for this type of parabola is a horizontal line of the form:

`y = k - f`

where k is the y coordinate of the vertex and f is the focal distance

The equation of the directrix is:

`y = -67/4`