Question

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Answer 1

Share how you managed that result, please.

Answer 2

yes "a" is indeed 12
so what's "b"? we dunno
but we know that the foci are at \(0, \pm 11\) thus

notice that the distance between the center of the ellipse and either foci is 11 units
so \(\bf \textit{distance from center to foci}=c=\sqrt{a^2-b^2}
\\ \quad \\
c^2=a^2-b^2\implies b=\sqrt{a^2-c^2}\qquad {\color{blue}{ a=12\qquad c=11}}\\ \quad \\\implies b=\sqrt{12^2-11^2}\)

Answer 3

well, notice the major axis is over the y-axis, meaning the "a" component goes under the fraction with the "y" in it thus \(\bf \cfrac{(x-h)^2}{b^2}+\cfrac{(y-k)^2}{a^2}=1\implies\cfrac{(x-0)^2}{23}+\cfrac{(y-0)^2}{144}=1
\\ \quad \\
\implies \cfrac{x^2}{23}+\cfrac{y^2}{144}=1\)

Answer 4

yw