Let the point of intersection of the line y=mx with the chord be (n,an) Then n=
4+x1
2
⇒x1=2n−4
an=
4+y1
2
⇒y1=2an−4 since (x1,y1) is on the parabola then, (2an−4)2=4(2n−4) 4a2n2+16−16an=8n−16 4a2n2−16an−8n+32=0 4a2n2−(16a+8)n+32=0 To have two distinct chords, D>0. (16a+8)2−4(4a2)(32)>0 64(2a+1)2−512a2>0 (2a+1)2−8a2>0 4a2−4a+1<2 (2a−1)2<2 −√2<(2a−1)<√2 $ {1-\√ {2}}/{2}