ધારો કે $(h, 0)\,\, X -$ અક્ષ પર આવેલ બિંદુ છે, તો $(h, 0)$ મધ્યબિંદુવાળી જીવાનું સમીકરણ $T = S_1$ થશે.

$xh\,\, – \,\,\frac{1}{2}\,\,p\,\,(x + h)\,\, – \,\,\frac{1}{2}\,\,q\,\,(y + 0)\,\, = \,\,{h^2} – \,\,ph\,$

તે $(p\,,\,\,q)$ માંથી પસાર થાય છે ,જેથી 

$ph – \frac{1}{2}\,p\,\,(p + h)\,\, – \,\,\frac{1}{2}\,\,q\,.\,q\,\, = \,\,{h^2} – \,\,ph$'

$ \Rightarrow \,\,ph – \frac{1}{2}\,\,{p^2} – \frac{1}{2}\,\,ph\,\, – \,\,\frac{1}{2}\,\,{q^2} = \,\,{h^2} – ph\,\,\,$

$ \Rightarrow \,\,{h^2} – \frac{3}{2}\,\,ph\,\, + \,\frac{1}{2}\,\,({p^2} + {q^2})\,\, = \,\,0\,$

$h$ વાસ્તવિક છે કે જેથી ${B^2} – 4AC\,\, > \,\,0$

$\,\,\frac{9}{4}\,\,{p^2} – \,\,4\,\,.\,\,\frac{1}{2}\,({p^2} + {q^2})\,\, > \,\,0$

$\,\,9{p^2} – 8\,\,({p^2} + {q^2})\,\, > \,\,0\,\,\,$

$ \Rightarrow \,\,{p^2} – 8{q^2}\, > \,\,0\,\, \Rightarrow \,\,{p^2}\, > \,\,8{q^2}$