What is the standard form of the equation of the parabola with a directrix at x=3 and a focus at (1,1)?

1 Answer
Feb 22, 2018

#y = sqrt(-4x+8) + 1# and #y = -sqrt(-4x+8) + 1#

Explanation:

When you see directrix, think of what that line means. When you draw a line segment at 90 degrees from directrix, that segment will meet your parabola. The length of that line is same as the distance between where your segment met your parabola and your focus point. Let's change this into math syntax:

"line segment at 90 degrees from directrix" means the line will be horizontal. Why? The directrix is vertical in this problem (x=3)!

"length of that line" means the distance from directrix to the parabola. Let's say that the a point on parabola has #(x,y)# coordinate. Then the length of that line would be #(3-x)_#.

"distance between where your segment met your parabola and your focus point" means the distance from #(x,y)# to your focus. That would be #sqrt((x-1)^2+(y-1)^2)# .

Now, "The length of that line is same as the distance between where your segment met your parabola and your focus point." So,

#sqrt((x-1)^2+(y-1)^2) = 3 - x#
#(x-1)^2+(y-1)^2 = (3-x)^2#
#x^2-2x+1+(y-1)^2 = 9 - 6x+x^2#
#(y-1)^2 = -4x + 8#
#y-1 = +-sqrt(-4x+8)#

#y = sqrt(-4x+8) + 1#
and
#y = -sqrt(-4x+8) + 1#

Does it surprise you that you have two equations for the parabola? Well look at the shape of the parabola and think about why there would be two equations. See how for every x, there are two y values?

graph{(y-1)^2 = -4x + 8 [-10.13, 9.87, -3.88, 6.12]}

Sorry, but I don't think you can make #y = ax^2+bx+c# format for this question.