Circle inscribed in a parabola, find points of contact, 急!!

Centre of circle = (0, a)
Radius = 1
therefore, equation of circle is x^2 + (y -a)^2 = 1...............(1)
equation of parabola is y = 2x^2..............(2)
Sub. (2) into (1) we get
y/2 + (y -a)^2 - 1 = 0
y/2 + y^2 + a^2 - 2ay - 1 = 0
y + 2y^2 + 2a^2 - 4ay - 2 = 0
2y^2 + (1 - 4a)y + (2a^2 - 2) = 0
Since the circle touches the parabola, delta = 0
(1-4a)^2 - 4(2)(2a^2 - 2) = 0
1 + 16a^2 - 8a - 16a^2 + 16 = 0
a = 17/8.
so y = -(1-4a)/4 = -(1-17/2)/4 = 15/8 (based on formula of quadratic equation with equal roots.)
so x = +/- sqrt(15/16). (based on the equation y= 2x^2)
so the point of contacts are : [(sqrt15)/4, 15/8] and [-sqrt15)/4, 15/8]

circle
x^2 + (y-a)^2 = 1 (1)
parabola
y= 2x^2 (2)

Put (2) in (1)

x^2 + (2x^2 - a) ^ 2 =1

x^2 + 4x^4 + a^2 - 4ax^2 = 1

4x^4 -3a x^2 + (a^2 -1 ) = 0

solution
x^2 = (3a +/- (9a^2 - 16a^2 -16)^(1/2)) / 8

x^2 must be + and two must only be two solution, so
the determinant must be 0, i.e. 9a^2 -16a^2 -16 =0. Otherwise, there would 4 solutions.

7a^2 =16

a^2 = 16/7

a= (16/7) ^ (1/2) since a must be positive, so -ve value is discarded.