Chapter 8.6, Problem 4GP

Solution

To Graph: the equation and identify the important characteristics of the graph.

Hence, Center of ellipse is $\left(2,1\right)$ , Vertices are at $\left(\pm 4,0\right)$ , Co-vertices are at $\left(0,\pm 3\right)$ and major axis is horizontal.

Given: An equation of ellipse is $\frac{{\left(x-2\right)}^2}{16}+\frac{{\left(y-1\right)}^2}{9}=1$

Concept Used: The standard equation of a parabola whose centre is at $\left(h,k\right)$ is $\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b}=1$.

Calculation:

Equation of hyperbola $\frac{{\left(x-2\right)}^2}{16}+\frac{{\left(y-1\right)}^2}{9}=1$

By comparing with standard equation of ellipse: center of ellipse is $\left(2,1\right)$.

Vertex of ellipse:

  $\begin{array}{l}{a}^2=16\\ a=\pm \sqrt{16}\\ a=\pm 4\end{array}$

Vertices of ellipse is $\left(\pm 4,0\right)$.

Co-vertices:

  $\begin{array}{l}{b}^2=9\\ b=\pm \sqrt{9}\\ b=\pm 3\end{array}$

Co-vertices of ellipse: $\left(0,\pm 3\right)$

Major axis is Horizontal.

Graphing of ellipse $\frac{{\left(x-2\right)}^2}{16}+\frac{{\left(y-1\right)}^2}{9}=1$:

  

Conclusion:

Hence, Center of ellipse is $\left(2,1\right)$ , Vertices are at $\left(\pm 4,0\right)$ , Co-vertices are at $\left(0,\pm 3\right)$ and major axis is horizontal.