Question 1128278
find the equation satisfied by the set of points such that each is equidistant from P(2,-1) and 3x+y-7=0
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It's a parabola.  The focus is (2,-1) and the directrix is 3x+y-7 = 0
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Find the equation of the axis of the parabola, the line thru (2,-1) and perpendicular to (call it L1) 3x+y-7=0.
Slope of L1 = -3
Slope of perpendicular line = 1/3
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y+1= (1/3)*(x-2)
The intersection of the 2 lines is D(13/5,-4/5)
The midpoint of PD is M(23/10,-9/10), a point on the parabola.
The distance from M to the line L1 is MD = PM = 
{{{sqrt((23/10-2)^2 + (-1 - 9/10)^2)}}}
= {{{sqrt(9/100 + 1/100) = sqrt(10)/10}}}
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The distance from a point (m,n) to a line Ax + By + C = 0 is:
(Am + Bn + C)/sqrt(A^2 + B^2)
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The distance from P to the line L is:
|2*3 -1*1 -7|/sqrt(9+1) = 2/sqrt(10)
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Line L2 parallel to L1 thru (2,-1) is y+1 = -3(x-2)
The 2 points on L2 at a distance of sqrt(10)/10 from P are Q(2.1,-1.3) and 
R(1.9,-0.7)
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Find the parabola thru the 3 points about the axis.
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More later.
My Excel sheet always finds a parabola with its axis parallel to the y-axis.
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Use this method:
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For any point (x,y) the distance to (2,-1) is sqrt((x-2)^2 + (y+1)^2)
The distance to the line L1 is |3x + y - 7|/sqrt(10)
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{{{d^2 = (x-2)^2 + (y+1)^2 = (3x+y-7)^2/10}}}
{{{10*(x^2-4x+4 + y^2+2y+1) = 9x^2-42x+6xy+y^2-14y+49}}}
{{{10x^2 - 40x + 40 + 10y^2 + 20y + 10 = 9x^2-42x+6xy+y^2-14y+49}}}
--> {{{x^2 + 2x + 1 + 9y^2 + 34y - 6xy = 0}}}