Is my solving method correct?

Question

Question: Let H: x2/a2 - y2/b2 = 1, a > 0, b > 0, be a hyperbola such that the sum of lengths of the transverse and the conjugate axes is 4(2√2 + √14). If the eccentricity H is √11/2, then the value of a2 + b2 is equal to ______.

Answer: Transverse axis, T.A.= 2a

              Conjugate axis, C.A.= 2b

              Therefore, Sum of T.A. and C.A.= 2a+2b= 2(a+ b)

              A/q, 2(a+b)= 4(2√2+√14)=> (a+b)=2(2√2+√14)=> (a+b)= (4√2+ 2√14)

              Since, x²+ y²= (x²+ y²)- 2xy

              Therefore, (a²+ b²)= (4√2+ 2√14)²- 2(4√2)(2√14)

              =>(a²+ b²)= (32+ 16√7+ 56)- 32√7

                               = 88+ 16√7- 32√7

              =>(a²+ b²)= 88

Answer 1

Correct answer is 88

\(\frac{x^2}{a^2}-\frac{y^2}{b^2}= 1\)

Given,

a² + b² = 32 + 56 = 88