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Using the definition of a parabola as the locus of points equidistant from a given point (focus) and a line (directrix).

Here is my work: $\sqrt{(y+x)^2}=\sqrt{(x-1)^2+(y-1)^2}$

$y^2+2xy+x^2=x^2-2x+1+y^2-2y+1$

$2xy+2y=-2x+2$

$y=\frac{-x+1}{x+1}$

This equation is a hyperbola not a parabola. Any suggestions on what I have done wrong would be appreciated.

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Let $(x,y)$ be a point on the parabola, its distance to the directrix is given by

$$\frac{|x+y|}{\sqrt{1^2+1^2}}$$

So the equation of the parabola is

$$(x-1)^2+(y-1)^2=\frac{(x+y)^2}{2}$$

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  • $\begingroup$ I hate to sound like an idiot, but I have no idea where you got this from. I thought I would use the distance formula and set distances equal. Could you please explain this to me. Thank You. $\endgroup$ – bob Jun 9 '17 at 2:29
  • $\begingroup$ I used the formula of the distance between a point and a line. See en.m.wikipedia.org/wiki/Distance_from_a_point_to_a_line. In your work, it seems that you take $\sqrt{(x+y)^2}$ as the distance between the point on the parabola and the directrix. This is not correct. $\endgroup$ – CY Aries Jun 9 '17 at 2:40
  • $\begingroup$ Ok thank you. I see where I was making a mistake. I have one last question. On the left hand side of your equation of the parabola, you have (y-2)^2 . I am not sure why it is y-2 and not y-1. Sorry to be such a bother. Thanks for all of your help. $\endgroup$ – bob Jun 9 '17 at 3:02
  • $\begingroup$ It's a typo. I'll correct it $\endgroup$ – CY Aries Jun 9 '17 at 3:03
  • $\begingroup$ Thanks again for all of your help today. $\endgroup$ – bob Jun 9 '17 at 3:04

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