Given
E≡3x2+4y2=12⇒4x2+3y2=1
Now, eE=1−a2b2=1−43=21
Length of L.R. =a2b2=22×3=3
Now H≡a2x2−1y2=1
Length of L.R.=a2b2=a2×1
Given length of L.R. of E and H are equal
So a2=3⇒a=32
Now eH=a2b2+1=(32)21+1⇒eH=49+1=413
So 12(eH2+eE2)=12(413+41)=12(414)=14×3=42