The correct option is C (94, 38)
2y dydx=(2−x)2 - 2x(2 - x), so dydx|(1,1)=12[1−2]=−12
Therefore, the equation of tangent at (1, 1) is y - 1 = -12(x−1)
⇒ 2y - 2 = -x + 1
⇒ y = −x+32
The intersection of the tangent and the curve is given by (14)(−x+3)2=x(4+x2−4x)
⇒x2−6x+9=16x+4x3−16x2
⇒4x3−17x2+22x−9=0
⇒(x−1)(4x2−13x+9)=0
⇒(x−1)2(4x−9)=0
Since x = 1 is already the point of tangency x = 94 and y2 = 94(2−94)2=964
Thus the required point is (94,38)