hyperbola centered at x =12/2 = 6
hits x axis at (5.5 , 0) and (6.5 , 0)
(x-6)^2/a^2 - y^2/b^2 = 1
2 a = 1
a = 1/2
a^2 = 1/4
4 (x-6)^2 - y^2/b^2 = 1
center to either focus = 4 =sqrt(a^2+b^2)
so
16 = 1/4 + b^2
b^2 = 16 - 1/4 = 63/4
4(x-6)^2 - 4y^2/63 = 1
(x-6)^2 - y^2/63 = 1/4
Find the equation of the graph of all points such that the difference of their distances from (10,0) and (2,0) is always 1.
1 answer
1 answer