Find the exact solution(s) of the system: (x^2/4)-y^2=1 and x=y^2+1

Question

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!!

Answer 1

(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