We know that the equation of normal for y2=4x is y=−xt+2t+t3 ....(1)
Since it intersects at x=2, we get y=t3
Let the three ordinates be t31,t22,t33 are in A.P.
2t32=t31+t33
⇒ =(t1+t3)3−3t1t3(t1+t3) ....(2)
Now, t1+t2+t3=0
⇒t1+t3=−t2
Hence, Eq.(2) reduces to
⇒2t32=(−t2)3−3t1t3(−t2)
⇒2t32=−t32+3t1t2t3
⇒3t32=3t1t2t3
⇒t22=t1t3
So, t1,t2,t3 are in G.P.
Hence, slopes of tangents 1t1, 1t2, 1t3 are in G.P.