Question

write the equation of an ellipse with minor axis endpoints (co - vertices) at (0, 3) and (0, - 3) and major axis endpoints (vertices) at (-11, 0) and (11, 0).
\\(\frac{x^{2}}{121}+\frac{y^{2}}{9}=1\\)
\\(\frac{x^{2}}{3}+\frac{y^{2}}{11}=1\\)
\\(\frac{x^{2}}{9}+\frac{y^{2}}{121}=1\\)
\\(\frac{x^{2}}{6}+\frac{y^{2}}{22}=1\\)

Explanation:

Step1: Determine the center of the ellipse

The center of the ellipse is the mid - point between the vertices or co - vertices. Since the mid - point between $(0,3)$ and $(0, - 3)$ is $(0,0)$ and the mid - point between $(-11,0)$ and $(11,0)$ is $(0,0)$, the center of the ellipse is $(h,k)=(0,0)$.

Step2: Find the values of $a$ and $b$

The length of the semi - major axis $a$ is the distance from the center to a vertex. The vertices are at $(-11,0)$ and $(11,0)$, so $a = 11$ and $a^{2}=121$. The length of the semi - minor axis $b$ is the distance from the center to a co - vertex. The co - vertices are at $(0,3)$ and $(0, - 3)$, so $b = 3$ and $b^{2}=9$.

Step3: Write the standard form of the ellipse equation

The standard form of the equation of an ellipse centered at the origin $(0,0)$ is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} = 1$ when the major axis is along the $x$ - axis. Substituting $a^{2}=121$ and $b^{2}=9$ into the equation, we get $\frac{x^{2}}{121}+\frac{y^{2}}{9}=1$.

Answer:

A. $\frac{x^{2}}{121}+\frac{y^{2}}{9}=1$