Answer:
no because it would be flipped
(y-k)²/a² + (x-h)²/b² = 1
Vertices being : A(0,1) & B(0, - 9), means that h=0
Let's calculate k: Af₁ + Af₂ = |2+8| = 2a = 10 → a=5 (f₁ and f₂ = focii)
1) a = 10 (WRONG) since a =5
2) h=3 (WRONG) since h =0
3) k =0 (WRONG) since k =5
4) b = 4 RIGHT; PROOF:
we know that : c² = a² - b²
c= 3 & a = 5 then → 3² = 5² - b²
9 = 25 - b²
b² = 25-9 = 16 and b = 4
