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#