The correct options are
C y=x−3a
D y=−2x+12a
Normal to y2=4ax is
y=−tx+at3+2at
Since it passes through (5a,2a)
∴2a=−5at+at3+2at⇒2a=at3−3at⇒t3−3t−2=0 (∵a≠0)
By observation t=−1 is one root,
⇒(t+1)(t2−t−2)=0⇒(t+1)2(t−2)=0∴t=−1 or t=2
So, the equation of normal are
t=−1⇒y=x−a−2a⇒y=x−3at=2⇒y=−2x+8a+4a⇒y=−2x+12a