Given, equation of parabola ⇒y2=2x Equation of normal of parabola, y2=4ax is tx+y=2at+at3 Here, 4a=2 So, a=1∕2 So, equation of normal ⇒tx+y=t+t3∕2 t3+(2−2x)t−2y=0 As, there are three normals which are passing through (a,0), so there must be three roots of this equation. t3+(2−2a)t−2.0=0 ⇒t3+(2−2a)t=0 ∴t1+t2+t3=−(2−2a)=2a−2 and t1t2+t2t3+t3t1=0 So, t12+t22+t32>0 (t1+t2+t3)2−2(t1t2+t2t3+t3t1)>0 ⇒(2a−2)2−2.0>0 ⇒a>1