It took me a little bit more playing before I understood your "axis lower on the OX axis". The book answer did help.

You probably meant that the major axis is along the y-axis.

So, with a = semi-major axis and b = semi-minor axis,

(x^2)/(b^2) +(y^2)/(a^2) = 1 -----(i)

a = 2b ----given

a^2 = 4b^2

At point (sqrt(7) /2,3),

Substitutions into (i),

[(Sqrt(7) /2)^2]/(b^2) + (3^2)/(4b^2) = 1

7/(4b^2) +9/(4b^2) = 1

Clear the fractions, multiply both sides by 4b^2,

7 + 9 = 4b^2

b = 2 units --------**

And so, a = 2b = 4 units ----**

Hence, the ellipse is

(x^2)/(2^2) +(y^2)/(4^2) = 1

(x^2)/4 +(y^2)/16 = 1

Clear the fractions, multiply both sides by 16,

4x^2 +y^2 = 16 -----------------------------answer.

The book answer is 16x^2 +4y^2 = 64.

The book answer is on steroids.