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=44=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)