Question

What is the standard form equation of the parabola with a directrix of x=5 and focus at (11, -7)?

Answer 1

Standard form is:

`x = 1/12y^2+14/12y+ 145/12`

Explanation:

Because the directrix is a vertical line, `x = 5`, the vertex form for the equation of the parabola is:

`x = 1/(4f)(y-k)^2+h" [1]"`

where (h,k) is the vertex and #f is the signed horizontal distance from the vertex to the focus.

We know that the y coordinate, k, of the vertex is the same as the y coordinate of the focus:

`k = -7`

Substitute -7 for k into equation [1]:

`x = 1/(4f)(y--7)^2+h" [2]"`

We know that the x coordinate of the vertex is the midpoint between the x coordinate of the focus and the x coordinate of the directrix:

`h= (x_"focus"+x_"directrix")/2`

`h= (11+5)/2`

`h = 16/2`

`h = 8`

Substitute 8 for h into equation [2]:

`x = 1/(4f)(y--7)^2+8" [3]"`

The focal distance is the signed horizontal distance from the vertex to the focus:

`f = x_"focus"-h`

`f = 11-8`

`f = 3`

Substitute 3 for f into equation [3]:

`x = 1/(4(3))(y--7)^2+8`

We will multiply the denominator and write -- as +

`x = 1/12(y+7)^2+8`

Expand the square:

`x = 1/12(y^2+14y+ 49)+8`

Distribute the `1/12`

`x = 1/12y^2+14/12y+ 49/12+8`

Combine the constant terms:

`x = 1/12y^2+14/12y+ 145/12`

Answer 2

`x=y^2/12+7/6y+145/12`

Explanation:

Directrix `x=5`
Focus `(11, -7)`
From this we can findout the vertex.
Look at the diagram

Vertex lies exactly in between Directrix and Focus

`x,y=(5+11)/2, (-7 + (-7))/2=(8, -7)`

The distance between Focus and vertex is `a=3`
The parabola is opening to the right
The equation of the Parabola here is -

`(y-k)^2=4a(x-h)`
`(h,k)` is the vertex
`h=8`
`k=-7`

Plugin `h=8; k=-7 and a=3` in the equation

`(y-(-7))^2=4.3(x-8)`
`(y+7)^2=4.3(x-8)`

`12x-96=y^2+14y+49` [by transpose]
`12x=y^2+14y+49+96`
`12x=y^2+14y+145`
`x=y^2/12+14/12y+145/12`
`x=y^2/12+7/6y+145/12`