)(2,4)=1 ∴ The equation of normal at the point (2,4) and slope −1, is y−4=−1(x−2) ⇒x+y−6=0 .......(ii) ∴ This line intersect the Eq. (i). Solving Eqs. (i) and (ii), we get (6−x)2=8x ⇒x2+36−12x=8x ⇒x2−20x+36=0 ⇒x=2,18 Point x=2 is already taken, now x=18⇒y=±12 Here, we take y=−12 because second point will be below the X -axis. ∴ Required point is (18,−12).