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Let O be the center of the ellipse and Let A'A = 2a be the major axis and B'B = 2b the minor axis.

Given: OA = a = 20;  OF₁ [focus = 16 , because its coordinates are (16,0)].
We know that from any point M on the ellipse we have the following equality:

MF₁+MF₂ = 2a (F₁ and F₂ being the foci).

When M is on B, we have the following equality: BF₁ + BF₂ =2a = 40
But BF₁ = BF₂ → BF₁ = 20
In the right triangle BOF₁ . Apply Pythagoras:BF1² = OB² + OF₁²
BF₁² = 400; OF₁² = 256 Then OB² = 400-256 = 144
OB = √144 , and OB = b = 12. ANSWER : b =12.

Having found b, the equation becomes:
x²/400 + y²/144 = 1