Question

question 11 · 1 point given the ellipse
\\(\frac{(x - 4)^2}{144}+\frac{(y + 3)^2}{16}=1\\)
what are the vertices and co - vertices of the ellipse?
select the correct answer below:
the vertices are at (4,9) and (4, - 15). the co - vertices are at (8, - 3) and (0, - 3).
the vertices are at (4,1) and (4, - 7). the co - vertices are at (16,1) and ( - 8, - 3).
the vertices are at (16, - 3) and ( - 8, - 3). the co - vertices are at (4,1) and (4, - 7).
the vertices are at (8, - 3) and (0, - 3). the co - vertices are at (4,9) and (4, - 15).

Explanation:

Step1: Identify the form of the ellipse equation

The standard form of an ellipse centered at $(h,k)$ is $\frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1$. For the given ellipse $\frac{(x - 4)^2}{144}+\frac{(y + 3)^2}{16}=1$, we have $h = 4,k=-3,a^2 = 144,a = 12,b^2=16,b = 4$.

Step2: Find the vertices

Since $a^2$ is under the $(x - h)^2$ term, the major axis is horizontal. The vertices are given by $(h\pm a,k)$. Substituting the values, we get $(4\pm12,-3)$, which are $(16,-3)$ and $(-8,-3)$.

Step3: Find the co - vertices

The co - vertices are given by $(h,k\pm b)$. Substituting the values, we get $(4,-3\pm4)$, which are $(4,1)$ and $(4,-7)$.

Answer:

C. The vertices are at $(16, - 3)$ and $(-8, - 3)$. The co - vertices are at $(4,1)$ and $(4, - 7)$.