Find the exact solution(s) of the system: (x^2/4)-y^2=1 and x=y^2+1
Answer:(4,sqrt3),(4,-sqrt3)
2)Write an equation for an ellipse if the endpoints of the major axis are at (-8,1) and (8,1) and the endpoints of the minor axis are at (0,-1) and (0,3).
Answer: (x^2/64) + (y-1)^2/4 = 1
Thanks a lot!!
(x^2/4)-y^2=1 and x=y^2+1
rewrite the second as y^2 = x-1 and sub into the first
x^2 /4 - (x-1) = 1
x^2 - 4x + 4 = 4 , I multiplied each term by 4
x(x-4) = 0
so x=0 or x=4
in y^2 = x-1
if x=0, y^2 = -1 ----> no solution
if x=4 , y^2 = 3 ---> y = ±√3
so they intersect at (4,√3) and (4,-√3)
2.
For the ellipse, the centre is (0,1), the midpoint of (-8,1) and (8,1)
(notice the (0,1) is also the midpoint of the minor axis)
a=8 and b=2, so....
(x^2)/64 + (y-1)^2 /4 = 1