Answer:
Given curves are
x2
= 4y ?. (1)
x = 4y ? 2 ?..(2)
For (1), x2
= 4y
(1) and (2) (4y ?
2)2 = 4y
16y2
+ 4 ? 16y = 4y
16y2
? 20y + 4 = 0
4y2
? 5y + 1 = 0
4y2
? 4y ? y + 1 = 0
4y (y ?
1) ? (y ? 1) = 0
y
= 1, ¼
x
= 2, ? 1 [by (2)]
Points
of intersection are (?1, ) and
(2, 1)
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