Chapter 8, Problem 11RE

Solution

To find: The center, vertices and foci of the identified conic $\frac{{\left(x-2\right)}^2}{16}+\frac{{\left(y+1\right)}^2}{7}=1$ and sketch its graph.

Ellipse. Center is $\left(2,-1\right)$ ,vertices: $\left(6,-1\right)$ and $\left(-2,-1\right)$ ; foci: $\left(5,-1\right)$ and $\left(-1,-1\right)$ .

Given information:

Equation of conic: $\frac{{\left(x-2\right)}^2}{16}+\frac{{\left(y+1\right)}^2}{7}=1$

Formula used:

The standard equation of the hyperbola is $\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$

Center is $\left(h,k\right)$

Foci is $\left(h\pm c,k\right)$

Vertices is $\left(h\pm a,k\right)$

Pythagorean relation: $c^2=a^2-b^2$

Calculation:

The equation $\frac{{\left(x-2\right)}^2}{16}+\frac{{\left(y+1\right)}^2}{7}=1$ is of the form of standard equation of the hyperbola $\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$

Substitute $2$ for $h$ and $-1$ for $k$ in the center formula.

Center is $\left(2,-1\right)$

Compare the equation with the standard equation of the hyperbola.

  $a^2=16$

Take square root on both sides of the equation.

  $b^2=7$

Substitute $16$ for $a^2$ and $7$ for $b^2$ in the Pythagorean relation.

  $\begin{array}{c}{c}^2=16-7\\ =9\end{array}$

Take square root on both sides of the equation.

  $\begin{array}{c}c=\sqrt{9}\\ =3\end{array}$

Substitute $2$ for $h$ , $3$ for $c$ and $-1$ for $k$ in the foci formula.

  $\begin{array}{l}\left(2\pm 3,-1\right)\\ \left(2+3,-1\right)=\left(5,-1\right)\\ \left(2-3,-1\right)=\left(-1,-1\right)\end{array}$ .

Substitute $2$ for $h$ and $-1$ for $k$ and $4$ for $a$ in the vertices formula.

  $\begin{array}{l}\left(2+4,-1\right)=\left(6,-1\right)\\ \left(2-4,-1\right)=\left(-2,-1\right)\end{array}$ .

Draw the graph as follows:

  

Hence, the given equation of conic is an ellipse, center is $\left(2,-1\right)$ ,vertices is $\left(6,-1\right)$ and $\left(-2,-1\right)$ and foci is $\left(5,-1\right)$ and $\left(-1,-1\right)$ .