Question 810701
<pre>
Find the equation of the parabola whose focus is (-3, 2) and directrix is x+y = 4.

{{{drawing(400,1000/3,-15,9,-10,10,
green(line(-9,13,16,-12)),
 graph(400,1000/3,-15,9,-10,10, x+sqrt(-20x-10)),
circle(-3,2,.1),circle(-3,2,.05),circle(-3,2,.15),circle(-3,2,.125),circle(-3,2,.025),locate(-5.5,1.7,"(-3,2)"),
graph(400,1000/3,-15,9,-10,10, x-sqrt(-20x-10)) )}}}

By definition a parabola is the set of points each of which is the same
distance from a given point (called the focus) as its perpendicular 
distance to a given line (called the directrix).

Let (x,y) be any arbitrary point on the parabola.
The green line is the directrix with equation x+y = 4.
Draw two lines, one from that arbitrary point (x,y) 
perpendicular to the directrix and another from (x,y)
to the focus (-3,2): 

{{{drawing(400,1000/3,-15,9,-10,10,
green(line(-9,13,16,-12)),
 graph(400,1000/3,-15,9,-10,10, x+sqrt(-20x-10)),
circle(-3,2,.1),circle(-3,2,.05),circle(-3,2,.15),circle(-3,2,.125),circle(-3,2,.025), locate(-5.5,1.7,"(-3,2)"),

circle(-9.5,3.916407865,.1),circle(-9.5,3.916407865,.05),circle(-9.5,3.916407865,.15),circle(-9.5,3.916407865,.125),circle(-9.5,3.916407865,.025),locate(-11.7,5.6,"(x,y)"),
line(-4.70820393,8.70820393,-9.5,3.916407865),
line(-3,2,-9.5,3.916407865),

graph(400,1000/3,-15,9,-10,10, x-sqrt(-20x-10)) )}}}

Those two black lines must be equal in length.

The perpendicular distance from the point (x<sub>1</sub>,y<sub>1</sub>)
to the line Ax+By+C=0 is
d = {{{abs((Ax[1]+By[1]+C)/sqrt(A^2+B^2))}}}

Therefore the perpendicular distance from any arbitrary point 
(x,y) on the desired parabola to the line x+y=4 or x+y-4=0 is

d = {{{abs((1x+1y-4)/sqrt(1^2+1^2))}}} = {{{abs(x+y-4)/sqrt(2))}}} = {{{abs((x+y-4)/sqrt(1+1))}}} = {{{abs(x+y-4)/sqrt(2))}}}

The distance between two points (x<sub>1</sub>,y<sub>1</sub>) and (x<sub>2</sub>,y<sub>2</sub>)

is given by the formula 

d = &#8730;<span style="text-decoration: overline">(x<sub>2</sub>-x<sub>1</sub>)²+(y<sub>2</sub>-y<sub>1</sub>)²</span>

So the distance from any arbitrary point (x,y) and (-3,2) is given by:

d = &#8730;<span style="text-decoration: overline">(-3-x)²+(2-y)²</span>

So we set the two distances equal:

{{{abs(x+y-4)/sqrt(2))}}} = &#8730;<span style="text-decoration: overline">(-3-x)²+(2-y)²</span>

Multiply both sides by &#8730;<span style="text-decoration: overline">2</span> 

|x+y-4| = &#8730;<span style="text-decoration: overline">2</span>&#8730;<span style="text-decoration: overline">(-3-x)²+(2-y)²</span>

Squaring both sides gets rid of the absolute value and the square roots:

(x+y-4)² = 2[(-3-x)²+(2-y)²]
x²+y²+16+2xy-8x-8y = 2[9+6x+x²+4-4y+y²]
x²+y²+16+2xy-8x-8y = 18+12x+2x²+8-8y+2y²
x²+y²+16+2xy-8x-8y = 26+12x+2x²-8y+2y²
 -x²+2xy-y²-20x-10 = 0
  x²-2xy+y²+20x+10 = 0

Edwin</pre>