JEE Main 24 Feb 2021 Shift 2 Solved Paper

Given, curve y=x2+4
and, line y=4x−1
Here, y=x2+4∴‌

dy

dx

‌‌=2x . . . (i)
‌ and ‌y‌‌=4x−1
‌

dy

dx

‌‌=4 . . . (ii)
Let the required point be P(x1,y1).
∴‌‌‌

dy

dx

|p=2x1 . . . (iii)
∵ Slopes will be equal.
∴‌‌2x1=4‌‌
[from Eqs. (ii) and (iii)]
⇒ 2x1=4 4=2
Now, the given point P(x1,y1) lies on curve y=x2+4,
∴‌y1=x12+4
⇒‌y1=22+4=8
Hence, required coordinate of P=(2,8)