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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

Hello!

the answer is:

the answer is:

why?

to solve the problem, we need to remember the quotient of power property, it's defined by the following relation:

if we have a quotienf of powers that have the same base, we need to keep the same base and subtract the exponent of the denominator power to the exponent of the numerator power.

so, we are given the expression:

then, calculating we have:

hence, the answer is:

have a nice day!