How do you write an equation of a parabola with directrix x+y=1 and focus (1,1)?

1 Answer
Sep 22, 2016

Answer:

#x^2+y^2-2xy-2x-2y+3=0# graph{x^2+y^2-2xy-2x-2y+3=0 [-10, 10, -5, 5]}

Explanation:

We use the Focus-Directrix Property of Parabola (FDP) :

(FDP) : Let pt. #S", and, line "d# be the Focus and Directrix of a

Parabola, resp. If #P# is any pt. on the parabola, then, the pt.#P# is

equidistant from the pt. #S# and the line #d#.

Recall that the #bot"-dist. of a pt."P(x_0,y_0)"# from a line :

#ax+by+c+0# is given by #|ax_0+by_0+c|/sqrt(a^2+b^2).#

Let #P(x,y)# be any pt. on the Parabola. Then, by the discussion

above, we have,

#sqrt{(x-1)^2+(y-1)^2}=|x+y-1|/sqrt(1^2+1^2)#. Squaring both sides,

#:. 2(x^2+y^2-2x-2y+2)=x^2+y^2+(-1)^2+2xy-2y-2x#, or,

#x^2+y^2-2xy-2x-2y+3=0#