Find the equation of the parabola whose focus is (-3, 2) and directrix is x+y = 4.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): Those two black lines must be equal in length. The perpendicular distance from the point (x1,y1) to the line Ax+By+C=0 is d = 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 = = = = The distance between two points (x1,y1) and (x2,y2) is given by the formula d = √(x2-x1)²+(y2-y1)² So the distance from any arbitrary point (x,y) and (-3,2) is given by: d = √(-3-x)²+(2-y)² So we set the two distances equal: = √(-3-x)²+(2-y)² Multiply both sides by √2 |x+y-4| = √2√(-3-x)²+(2-y)² 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