Getting nowhere , plugging values seems better
s²+c²=1 so assume s=1/√ 3 and c =√ (2/3 ) have to choose at diff. values as possible
c=s/t so t = s/c = 1/√ t
y=bt so if b =√ 2, y = 1 better to have the value of y as 1
x=as so if a = 3 , x = √ 3
(A) y √ (a^2 + x^2) = b x => y √ (a^2 + x^2) = 1√(9+3) but bx = √6
(B) y √ (a^2 - x^2) = b x => y √ (9-3) = bx = √6
(C) y √ (x^2 - a^2) = b x
(D) y √ (b^2 - x^2) = a x
(E) y √ (b^2 + x^2) = a x=> √(2+3) ,ax =3 √3
The eq are pretty

IMO B