Question

question 8 · 1 point given the ellipse $\frac{x^{2}}{36}+y^{2}=1$ what are the vertices and co - vertices of the ellipse? select the correct answer below: the vertices are at (1,0) and (-1,0). the co - vertices are at (0,6) and (0, - 6). the vertices are at (6,0) and (-6,0). the co - vertices are at (0,1) and (0, - 1). the vertices are at (0,1) and (0, - 1). the co - vertices are at (6,0) and (-6,0). the vertices are at (0,6) and (0, - 6). the co - vertices are at (1,0) and (-1,0).

Explanation:

Step1: Recall the standard form of an ellipse

The standard form of a horizontal ellipse is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} = 1$ ($a>b>0$), where the vertices are $(a,0)$ and $(-a,0)$ and co - vertices are $(0,b)$ and $(0,-b)$. Given $\frac{x^{2}}{36}+y^{2}=1$, we can rewrite it as $\frac{x^{2}}{6^{2}}+\frac{y^{2}}{1^{2}} = 1$.

Step2: Identify $a$ and $b$ values

Here, $a = 6$ and $b = 1$.

Step3: Determine vertices and co - vertices

Vertices: Since $a = 6$, vertices are $(6,0)$ and $(-6,0)$. Co - vertices: Since $b = 1$, co - vertices are $(0,1)$ and $(0,-1)$.

Answer:

B. The vertices are at $(6,0)$ and $(-6,0)$. The co - vertices are at $(0,1)$ and $(0, - 1)$.