Will give medal ! and points
what are the vertices foci and asymptotes of the hyperbola? is the hyperbola horizontal or vertical? 14y^2-28x^2=448

Hyperbola equation:

(x-h)² /a² +  (y-k)² /b²=1 if the major axis is a
and (y-k)²/b² +(x-h)²/a² 1 if the major axis is b
Let's solve your equation:

14y²-28x²=448 , Divide all part by 448:
14y²/448 - 28x²/448 =448/448
After simplification y²/32 - x²/16 =1 with h=0 & k=0. that means its
centre is at (0,0)

Since b (b²=32 & b=4√2) is the major axis (b=4√2 & a= 4), hence it's a vertical hyperbola (open up & down)

Vertices since h=k=0, the vertices are at +b & -b that is +4√2 & -√2 so there coordinates are (0, 5.657) & (0, - 5.657)

Foci = √(a²+b²)  & - √(a²+b²)
F₁ = √(32+16) & F₂ = - √(32+16)==>F₁=6.928 & F₂= -6.928

Asymptotes = +b/a(x) & -b/a(x),

Asymptote₁=>  y₁=(√32)/4 (x)  ==> y₁= 1.414 (x)

Asymptote₂=>  y₂= -(√32)/4 (x) ==> y₂=-1.414 (x)

The answer would be a i think