Algebra II
posted by Lucy .
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) + (y1)^2/4 = 1
Thanks a lot!!

(x^2/4)y^2=1 and x=y^2+1
rewrite the second as y^2 = x1 and sub into the first
x^2 /4  (x1) = 1
x^2  4x + 4 = 4 , I multiplied each term by 4
x(x4) = 0
so x=0 or x=4
in y^2 = x1
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 + (y1)^2 /4 = 1
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