Let the two parabolas be
y2=4ax.......(i)y2=−4ax......(ii)
Let tangents are drawn from a point P(at2,2at) on (i) to (ii)
Then QQ′ is a chord of contact to (ii) w.r.t. to P
Equation of chord of contact is T=0
yy1=−2a(x+x1)2aty=−2a(x+at2)ty=−x−at2......(iii)
substituting x from (iii) in (i)
y2=4a(−at2−ty)y2+4aty+4a2t2=0(y+2at)2=0⇒y=−2at
We are getting equal roots from the equation. So, QQ′ touch the parabola
Substituting y=−2at in (i)
(−2at)2=4ax⇒x=at2
So the point of contact to (i) is P′(at2,−2at)
Equation of common tangent at O is
x=0
Slope of PP′=−2at−2atat2−at2=∞
So, PP′ is perpendicular to x axis
Hence it is parallel to the y axis or x=0
Hence proved.