Question

The general form of an parabola is 2x2−12x−3y+12=0 .

What is the standard form of the parabola?

Answer 1

I took the test

Explanation:

Answer 2

The standard form of the parabola is
`(x-3)^2=4*(3)/(8)(y+2)`

Explanation:

The standard form of a parabola is

`(x-h)^2=4p(y-k)`.

In order to convert
`2x^2-12x-3y+12=0` into the standard form, we first separate the variables:

`2x^2-12x+3y+12=0\\\\2x^2-12x+12=3y`

we now divided both sides by 2 to remove the coefficient from
`2x^2` and get:

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

We complete the square on the left side by adding 3 to both sides:

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

`x^2-6x+9=(3)/(2)y+3`

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

now we bring the right side into the form
`4p(y-k)` by first multiplying the equation by
`(2)/(3)`:

`(2)/(3) *(x-3)^2=(2)/(3) *((3)/(2)y+3)\\\\(2)/(3) *(x-3)^2=y+2`

and then we multiplying both sides by
`(3)/(2)` to get

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

Here we see that

`4p=(3)/(2)`

`\therefore p=(3)/(8)`

Thus, finally we have the equation of the parabola in the standard form:

`\boxed{(x-3)^2=4*(3)/(8)(y+2)}`