A set of points P(x,y) has the property that the distance from (-2,0) minus the distance from (4,0) is 4. Derive the equation of the conic (using distances) and verify that it is a hyperbola. Solution Distance from (-2,0) - distance from (4,0) = 4 [(x+2)2+y2] - [(x-4)2+y2] = 4 [(x+2)2+y2] = 4 + [(x-4)2+y2] (x+2)2+y2 = 16 + (x-4)2+y2 + 8[(x-4)2+y2] x2+y2+4+4x = 16 +16 + x2 - 8x + y2 +8[(x-4)2+y2] 12x - 28 = 8[(x-4)2+y2] 3x - 7 = 2[(x-4)2+y2] 9x2+49-42x = 4[x2+y2+16-8x] 5x2-4y2-10x-15 = 0 Hence the equation is 5x2-4y2-10x-15 = 0 Comparint it with Ax2+Bxy+Cy2+Dx+Ey+F=0 A=5;B=0;C=-4,D=-10,E=0;F=-15 Eccentricity === [(2{(A-C)2+B2})/({A+C}+{(A-C)2+B2})] As B=0 here the formula reduces to = [2(A-C)/A+C+A-C] =[(A-C)/A] = 9/5 > 1 Hence the equation obtained is hyperbola.

1. A set of points P(x,y) has the property that the distance from (-2,0) minus the distance from (4,0)
is 4. Derive the equation of the conic (using distances) and verify that it is a hyperbola.
Solution
Distance from (-2,0) - distance from (4,0) = 4
[(x+2)2+y2] - [(x-4)2+y2] = 4
[(x+2)2+y2] = 4 + [(x-4)2+y2]
(x+2)2+y2 = 16 + (x-4)2+y2 + 8[(x-4)2+y2]
x2+y2+4+4x = 16 +16 + x2 - 8x + y2 +8[(x-4)2+y2]
12x - 28 = 8[(x-4)2+y2]
3x - 7 = 2[(x-4)2+y2]
9x2+49-42x = 4[x2+y2+16-8x]
5x2-4y2-10x-15 = 0
Hence the equation is
5x2-4y2-10x-15 = 0
Comparint it with
Ax2+Bxy+Cy2+Dx+Ey+F=0
A=5;B=0;C=-4,D=-10,E=0;F=-15
Eccentricity === [(2{(A-C)2+B2})/({A+C}+{(A-C)2+B2})]
As B=0 here
the formula reduces to
= [2(A-C)/A+C+A-C]
=[(A-C)/A]
= 9/5
> 1
Hence the equation obtained is hyperbola