Given, y=−√x+2 ......(i) And bisector of first quadrant is y=x .......(ii) On solving Eqs. (i) and (ii), we get x=1,4 ∴ From Eq. (i) Points are (1,1) and (4,0) But (4,0) not satisfy Eq. (ii) ∴ Point (1,1) is only point of intersection of curve (i) and line (ii). Now, slope of tangent of curve (i) is
dy
dx
=
−1
2√x
∴ Slope of tangent at point (1,1) is
dy
dx
|(1,1)=
−1
2
∴ Slope of normal at point (1,1) =
−1
dy
dx
|(1,1)
=
−1
−
1
2
=2 ∴ Equation of the normal to the curve at point (1,1) will be y−1=2(x−1)⇒2x−y−1=0