equation of a parabola with a weird axis

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October 18th 2008, 05:25 PM
HTB

equation of a parabola with a weird axis

I need to find the equation of a parabola with focus (-1,1) and directrix at y=x-2

I know what it looks like (kinda diagonal in the 2nd quadrant with vertex at the origin, but I dont know how to find its equation
October 18th 2008, 06:50 PM
Shyam

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Let P(x, y) be any point on the parabola.

Distance between P(x, y) and focus (-1, 1) = perpendicular distance from P(x, y) to directrix (x - y - 2 = 0)

$\sqrt{(x+1)^2 + (y-1)^2}= \left|\frac{x-y-2}{\sqrt{(-1)^2+1^2}}\right|$

$(x+1)^2 + (y-1)^2 = \left[\frac{x-y-2}{\sqrt{2}}\right]^2$

$x^2+y^2+8x-4y+2xy=0$

Please see attached diagram. Did you get it now???
October 22nd 2008, 01:27 AM
HallsofIvy

Here's another way to do it that might be simpler. Set up a new coordinate system, say x', y' such that the x' axis is parallel to y= x-2 and the y' axis is perpendicular to it: the line y= -x is perpendicular to y= x-2 and contains (-1,1) so that's the y' axis. y= -x and y= x-2 intersect at (1, -1) so the parabola itself goes through the point half way between (-1, 1) and (1, -1) which is, conveniently, (0, 0). The distance from the focus to the vertex of the parabola, (0,0), is sqrt(2) so the equation of the parabola in the x'y' coordinate system is y'= (1/4sqrt(2))x'. Now convert back to the xy coordinate system: if x' is rotated theta degrees from x, x'= x cos(theta)+ y sin(theta) and y= -x sin(theta)+ y cos(theta). Replace x' and y' in the formula with those.

Since the slope of y= x- 2 is 1, theta is 45 degrees and cos(theta) and sin(theta) are both sqrt(2)/2.