Chapter 8, Problem 9RE

Solution

To find: The center, vertices and foci of the identified conic $\frac{{\left(x+3\right)}^2}{18}-\frac{{\left(y-5\right)}^2}{28}=1$ and sketch its graph.

Hyperbola; center: $\left(-3,5\right)$ ; vertices: $\left(-3\pm 3\sqrt{2},5\right)$ ; foci: $\left(-3\pm \sqrt{46},5\right)$ .

Given information:

Equation of conic: $\frac{{\left(x+3\right)}^2}{18}-\frac{{\left(y-5\right)}^2}{28}=1$

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 $\frac{{\left(x+3\right)}^2}{18}-\frac{{\left(y-5\right)}^2}{28}=1$ 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 $-3$ for $h$ and $5$ for $k$ in the center formula.

Center is $\left(-3,5\right)$

Compare the equation with the standard equation of the hyperbola.

  $a^2=18$

Take square root on both sides of the equation.

  $\begin{array}{l}a=\sqrt{18}\\ a=\sqrt{9\times 2}\\ a=3\sqrt{2}\\ b^2=28\end{array}$

Take square root on both sides of the equation.

  $\begin{array}{c}b=\sqrt{28}\\ =\sqrt{7\times 4}\\ =2\sqrt{7}\end{array}$

Substitute $18$ for $a^2$ and $28$ for $b^2$ in the Pythagorean relation.

  $\begin{array}{c}{c}^2=28+18\\ =46\end{array}$

Take square root on both sides of the equation.

  $c=\sqrt{46}$

Substitute $-3$ for $h$ and $5$ for $k$ , $\sqrt{46}$ for $c$ in the foci formula.

  $\left(-3\pm \sqrt{46},5\right)$ .

Substitute $-3$ for $h$ and $5$ for $k$ , and $3\sqrt{2}$ for $a$ in the vertices formula.

Therefore, vertices are $\left(-3\pm 3\sqrt{2},5\right)$ .

Draw the graph as follows:

  

Hence, the given equation of conic is a hyperbola, center is $\left(-3,5\right)$ ,vertices is $\left(-3\pm 3\sqrt{2},5\right)$ and foci is $\left(-3\pm \sqrt{46},5\right)$ .