Q58 / 90 · JEE Main 2022 (24 Jun Shift 2) · Mathematics

Let the hyperbola H:x2a2-y2=1 and the ellipse E:3x2+4y2=12 be such that the length of latus rectum of H is equal to the length of latus rectum of E. If eH and eE are the eccentricities of H and E respectively, then the value of 12eH2+eE2 is equal to _____.

Correct answer: 42

Solution

Given

E≡3x2+4y2=12  ⇒ x24+y23=1

Now,  eE=1-b2a2=1-34=12

Length of L.R. =2b2a=2×32=3

Now H≡  x2a2-y21=1

 Length of L.R.=2b2a=2×1a

Given length of L.R. of E and H are equal

So 2a=3  ⇒  a=23

Now eH=b2a2+1=1232+1  ⇒  eH=94+1=134

So 12eH2+eE2=12134+14=12144=14×3=42

Hence, the correct answer is 42.

Relevant Questions