The correct option is A 4x+y=18
y=√x⇒x,y>0
Now squaring both the sides we get
y2=x
Any line orthogonal to parabolic curve is normal.
So, equation of normal is y=−tx+at3+2at
Here, a=14
Normal passes through (3,6)
⇒6=−3t+t34+2t4⇒24+12t=t3+2t⇒t3−10t−24=0
By observation, t=4 is one solution,
⇒(t−4)(t2+4t+6)=0
So, t=4 is the only real solution
∴ The required equation of normal is
y=−4x+644+84∴y+4x=18