Question

How do you write the parabola `2y^2+4y+x-8=0` in standard form and find the vertex, focus, and directrix?

Answer 1

You try to make a perfect square!

Explanation:

`2(y^2 +2y +1 -1) +x -8 = 0`

`2(y+1)^2-2+x-8=0`

`2(y+1)^2+x-10=0`

`2(y+1)^2 = -(x-10)`

`(y+1)^2 = -1/2(x-10)`

`(y+1)^2 = 4(-1/8)(x-10)`

Answer 2

This is a parabola that opens to the left, therefore, its standard form is:

`x = ay^2+by+c , a < 0`

The formulas for the requested items will be given in the explanation.

Explanation:

Given: `2y^2+4y+x-8=0`

To write in standard form, we add `-2y^2-4y+8` to both sides:

`x = -2y^2-4y+8`

The above is standard form where `a = -2`, `b = -4`, and `c = 8`

The vertex is a point `(h,k)`

The formula for the y-coordinate of the vertex is:

`k = -b/(2a)`

Substitute in the values, `a = -2` and `b = -4`

`k = -(-4)/(2(-2)`

`k = -1`

The formula for the x-coordinate of the vertex is:

`h = ak^2+b(k) + c`

Substitute in the values, `h = -1`, `a = -2`, `b = -4`, and `c = 8`

`h = -2(-1)^2-4(-1) + 8`

`h =10`

The vertex is `(10,-1)`

The focal distance `f` is:

`f = 1/(4a)`

`f = 1/(4(-2))`

`f = -1/8`

The formula for the focus is:

`(h+f, k)`

Substitute in values `h = 10`, `f = -1/8`, and `k = -1`

`(10-1/8, 10)`

The focus is `(79/8, -1)`

The formula for the equation of the directrix is:

`x = h-f`

Substitute in values `h = 10` and `f = -1/8`:

`x = 10 - -1/8`

`x = 81/8`

The above is the equation of the directrix.

The following is a drawing of the parabola, the vertex, the focus, and the directrix:

![www.desmos.com/calculator]()