∵(α,β) lies on the given ellipse, 25α2+4β2=1...(i) Tangent to the parabola, y=mx+
1
m
passes through (α,β). So, αm2−βm+1=0 has roots m1 and 4m1 m1+4m1=
β
α
and m1⋅4m1=
1
α
Gives that 4β2=25α...(ii) from (i) and (ii) 25(α2+α)=1...(iii) Now, (10α+5)2+(16β2+50)2 =25(2α+1)2+2500(2α+1)2 =2525(4α2+4α+1) from equation (iii) =2525(