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

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Answer 1 · 2016-09-22T14:09:25

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

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,

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