Question

How do you write the parabola `x^2-10x+12y+1=0` in standard form and find the vertex, focus, and directrix?

Answer 1

The standard Cartesian form for a parabola of this type is:

`y = ax^2+bx+c" [1]"`

Let's begin the conversion of the given equation:

`x^2-10x+12y+1=0`

to the form of equation [1] by subtracting `12y` from both sides:

`x^2-10x+1=-12y`

Multiply both sides of the equation by `-1/12`:

`-1/12x^2+10/12x-1/12=y`

Flip the equation and reduce the middle term:

`y = -1/12x^2+5/6x-1/12" [2]"`

Equation [2] is in the standard form of equation [1] where `a = -1/12, b =5/6 and c = -1/12`

The x coordinate of the vertex, h, can be found using the following equation:

`h = -b/(2a)`

Substitute in the values:

`h = -(5/6)/(2(-1/12))`

`h = -(5/6)/(-2/12)`

`h = -(5/6)-12/2`

`h = 5`

The y coordinate of the vertex, k, can be found by evaluating the equation at `x = h`:

`k = -1/12(5)^2+5/6(5)-1/12`

`k = 2`

The vertex is (5,2)

Find the focal distance, f, using the equation:

`f = 1/(4a)`

Substitute in the value of a:

`f = 1/(4(-1/12))`

`f = -12/4`

`f = -3`

The focus is the point described by the following pattern:

`(h,k+f)`

`(5,2-3)`

The focus is: `(5,-1)`

The equation of the directrix is the following:

`y = k-f`

`y = 2- (-3)`

The directrix is: `y = 5`