AMU Engineering 2019 solved paper Solved Paper

Given equation is y2=8x ...(i)
2y

dy

dx

=8⇒

dy

dx

=

4

y

⇒(

dy

dx

)(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).