need help with locus........ (1 Viewer)

[joto]

use the defination of a parabola and perpendicular distance formaula to find the equation og the parabola with focus S(-1,1) and directrix y=x-2

 

[LightXT]

CBF doing this myself.http://www.wolframalpha.com/input/?i=equation+of+the+parabola+with+focus+S(-1,1)+and+directrix+y=x-2
But I assume you'll want working.

 

[LightXT]

You have to solve this equation.
root((x+1)^2+(y-1)^2)=abs(x-y-2)/root(2)
http://www.wolframalpha.com/input/?i=root%28%28x%2B1%29%5E2%2B%28y-1%29%5E2%29%3Dabs%28x-y-2%29%2Froot%282%29
The LHS root((x+1)^2+(y-1)^2) gives the distance from a point to the focus S(-1,1).
The RHS abs(x-y-2)/root(2) gives the perpendicular distance from a point to the line y=x-2. You have to equate both sides. Sorry if my explanation is horrible.

 

[Nooblet94]

<a href="http://www.codecogs.com/eqnedit.php?latex=\textrm{Let}~P(x,y)~\textrm{represent any point on the parabola with focus}~S(-1,1)~\textrm{and directrix}~D~\textrm{given by the equation}~x-y-2=0.\\ ~\\ SP^2=(x@plus;1)^2@plus;(y-1)^2\\ DP=\frac{|x-y-2|}{\sqrt{1^2@plus;(-1)^2}}\\ DP^2=\frac{(x-y-2)^2}{2}\\ ~\\ \textrm{From the definition of a parabola, the distance from the focus to the parabola (SP) is equal to the distance from the directrix to the parabola (DP)}\\ ~\\ \therefore SP^2=DP^2\\ (x@plus;1)^2@plus;(y-1)^2=\frac{(x-y-2)^2}{2}\\ 2x^2@plus;4x@plus;2@plus;2y^2-4y@plus;2=x^2-2xy-4x@plus;y^2@plus;4y@plus;4\\ x^2@plus;8x@plus;2xy@plus;y^2-8y=0\\ (x@plus;y)^2@plus;8(x-y)=0" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\textrm{Let}~P(x,y)~\textrm{represent any point on the parabola with focus}~S(-1,1)~\textrm{and directrix}~D~\textrm{given by the equation}~x-y-2=0.\\ ~\\ SP^2=(x+1)^2+(y-1)^2\\ DP=\frac{|x-y-2|}{\sqrt{1^2+(-1)^2}}\\ DP^2=\frac{(x-y-2)^2}{2}\\ ~\\ \textrm{From the definition of a parabola, the distance from the focus to the parabola (SP) is equal to the distance from the directrix to the parabola (DP)}\\ ~\\ \therefore SP^2=DP^2\\ (x+1)^2+(y-1)^2=\frac{(x-y-2)^2}{2}\\ 2x^2+4x+2+2y^2-4y+2=x^2-2xy-4x+y^2+4y+4\\ x^2+8x+2xy+y^2-8y=0\\ (x+y)^2+8(x-y)=0" title="\textrm{Let}~P(x,y)~\textrm{represent any point on the parabola with focus}~S(-1,1)~\textrm{and directrix}~D~\textrm{given by the equation}~x-y-2=0.\\ ~\\ SP^2=(x+1)^2+(y-1)^2\\ DP=\frac{|x-y-2|}{\sqrt{1^2+(-1)^2}}\\ DP^2=\frac{(x-y-2)^2}{2}\\ ~\\ \textrm{From the definition of a parabola, the distance from the focus to the parabola (SP) is equal to the distance from the directrix to the parabola (DP)}\\ ~\\ \therefore SP^2=DP^2\\ (x+1)^2+(y-1)^2=\frac{(x-y-2)^2}{2}\\ 2x^2+4x+2+2y^2-4y+2=x^2-2xy-4x+y^2+4y+4\\ x^2+8x+2xy+y^2-8y=0\\ (x+y)^2+8(x-y)=0" /></a>

EDIT: last line had a - where there should have been a +

 

[LightXT]

^What program do you use for setting it out in such a clear manner?