(x−at12) ⇒‌‌(t2+t1)(y−2at1)‌=2(x−at12) ⇒‌‌(t1+t2)y−2x‌=2at1t2 . . . (i) This line is passing through (a,0) ⇒‌‌(t1+t2)(−2at1)=‌
2
t1+t2
(a−at12) ⇒‌‌t1t2=−1 . . . (ii) Let P(x1,y1) be the pole of (i) w.r.t. y2=4ax Its polar is yy1=2a(x+x1) . . . (iii) From equaiton (i) and (iii), we get ‌
t1+t2
y1
=‌
1
a
=‌
2at1t2
2ax1
From last two relations, we get ‌x1=at1t2 ⇒x1=−a ∴‌ locus is ‌x=−a‌