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

2 Answers
May 3, 2017

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

May 3, 2017

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
enter image source here

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