Chapter 11.3, Problem 47E

a.

To determine

To show: that the hyperbolas are confocal.

Explanation of Solution

Given information:

The hyperbola equation is $\frac{y^2}{k}-\frac{x^2}{16-k}=1$ .

  $0<k<16$ .

Concept used :if hyperbolas have same foci, so they are confocal.

prove: first find the foci of hyperbolas,

  $\frac{y^2}{k}-\frac{x^2}{16-k}=1$

  $\begin{array}{l}\frac{y^2}{1}-\frac{x^2}{16-1}=1\text{ [substitute k=1]}\\ y^2-\frac{x^2}{15}=1\end{array}$

Foci :

  $\begin{array}{l}c=\pm \sqrt{a^2+b^2}\\ c=\pm \sqrt{{\left(1\right)}^2+{\left(\sqrt{15}\right)}^2}\text{ [substitute a by 1 and b by }\sqrt{15}]\\ c=\pm \sqrt{16}\text{ [square and add]}\\ c=\pm 4\text{ [square root]}\end{array}$

Foci: $c\left(0,\pm 4\right)$

  $\frac{y^2}{k}-\frac{x^2}{16-k}=1$

  $\begin{array}{l}\frac{y^2}{2}-\frac{x^2}{16-2}=1\text{ [substitute k=2]}\\ \frac{y^2}{2}-\frac{x^2}{14}=1\end{array}$

Foci :

  $\begin{array}{l}c=\pm \sqrt{a^2+b^2}\\ c=\pm \sqrt{2+14}{\text{ [substitute a}}^2{\text{ by 2 and b}}^2\text{ by }14]\\ c=\pm \sqrt{16}\text{ [square and add]}\\ c=\pm 4\text{ [square root]}\end{array}$

Foci: $c\left(0,\pm 4\right)$

  $\frac{y^2}{k}-\frac{x^2}{16-k}=1$

  $\begin{array}{l}\frac{y^2}{3}-\frac{x^2}{16-3}=1\text{ [substitute k=3]}\\ \frac{y^2}{3}-\frac{x^2}{13}=1\end{array}$

Foci :

  $\begin{array}{l}c=\pm \sqrt{a^2+b^2}\\ c=\pm \sqrt{3+13}{\text{ [substitute a}}^2{\text{ by 3 and b}}^2\text{ by }13]\\ c=\pm \sqrt{16}\text{ [square and add]}\\ c=\pm 4\text{ [square root]}\end{array}$

Foci: $c\left(0,\pm 4\right)$

  $\frac{y^2}{k}-\frac{x^2}{16-k}=1$

  $\begin{array}{l}\frac{y^2}{4}-\frac{x^2}{16-4}=1\text{ [substitute k=4]}\\ \frac{y^2}{4}-\frac{x^2}{12}=1\end{array}$

Foci :

  $\begin{array}{l}c=\pm \sqrt{a^2+b^2}\\ c=\pm \sqrt{4+12}{\text{ [substitute a}}^2{\text{ by 4 and b}}^2\text{ by 12}]\\ c=\pm \sqrt{16}\text{ [square and add]}\\ c=\pm 4\text{ [square root]}\end{array}$

Foci: $c\left(0,\pm 4\right)$

  $\frac{y^2}{k}-\frac{x^2}{16-k}=1$

  $\begin{array}{l}\frac{y^2}{5}-\frac{x^2}{16-5}=1\text{ [substitute k=5]}\\ \frac{y^2}{5}-\frac{x^2}{11}=1\end{array}$

Foci :

  $\begin{array}{l}c=\pm \sqrt{a^2+b^2}\\ c=\pm \sqrt{5+11}{\text{ [substitute a}}^2{\text{ by 5 and b}}^2\text{ by }11]\\ c=\pm \sqrt{16}\text{ [square and add]}\\ c=\pm 4\text{ [square root]}\end{array}$

Foci: $c\left(0,\pm 4\right)$

  $\frac{y^2}{k}-\frac{x^2}{16-k}=1$

  $\begin{array}{l}\frac{y^2}{6}-\frac{x^2}{16-6}=1\text{ [substitute k=6]}\\ \frac{y^2}{6}-\frac{x^2}{10}=1\end{array}$

Foci :

  $\begin{array}{l}c=\pm \sqrt{a^2+b^2}\\ c=\pm \sqrt{6+10}{\text{ [substitute a}}^2{\text{ by 6 and b}}^2\text{ by }10]\\ c=\pm \sqrt{16}\text{ [square and add]}\\ c=\pm 4\text{ [square root]}\end{array}$

Foci: $c\left(0,\pm 4\right)$

  $\frac{y^2}{k}-\frac{x^2}{16-k}=1$

  $\begin{array}{l}\frac{y^2}{7}-\frac{x^2}{16-7}=1\text{ [substitute k=7]}\\ \frac{y^2}{7}-\frac{x^2}{9}=1\end{array}$

Foci :

  $\begin{array}{l}c=\pm \sqrt{a^2+b^2}\\ c=\pm \sqrt{7+9}{\text{ [substitute a}}^2{\text{ by 7 and b}}^2\text{ by 9}]\\ c=\pm \sqrt{16}\text{ [square and add]}\\ c=\pm 4\text{ [square root]}\end{array}$

Foci: $c\left(0,\pm 4\right)$

  $\frac{y^2}{k}-\frac{x^2}{16-k}=1$

  $\begin{array}{l}\frac{y^2}{8}-\frac{x^2}{16-8}=1\text{ [substitute k=8]}\\ \frac{y^2}{8}-\frac{x^2}{8}=1\end{array}$

Foci :

  $\begin{array}{l}c=\pm \sqrt{a^2+b^2}\\ c=\pm \sqrt{8+8}{\text{ [substitute a}}^2{\text{ by 8 and b}}^2\text{ by 8}]\\ c=\pm \sqrt{16}\text{ [square and add]}\\ c=\pm 4\text{ [square root]}\end{array}$

Foci: $c\left(0,\pm 4\right)$

  $\frac{y^2}{k}-\frac{x^2}{16-k}=1$

  $\begin{array}{l}\frac{y^2}{9}-\frac{x^2}{16-9}=1\text{ [substitute k=9]}\\ \frac{y^2}{9}-\frac{x^2}{7}=1\end{array}$

Foci :

  $\begin{array}{l}c=\pm \sqrt{a^2+b^2}\\ c=\pm \sqrt{9+7}{\text{ [substitute a}}^2{\text{ by 9 and b}}^2\text{ by 7}]\\ c=\pm \sqrt{16}\text{ [square and add]}\\ c=\pm 4\text{ [square root]}\end{array}$

Foci: $c\left(0,\pm 4\right)$

  $\frac{y^2}{k}-\frac{x^2}{16-k}=1$

  $\begin{array}{l}\frac{y^2}{10}-\frac{x^2}{16-10}=1\text{ [substitute k=10]}\\ \frac{y^2}{10}-\frac{x^2}{6}=1\end{array}$

Foci :

  $\begin{array}{l}c=\pm \sqrt{a^2+b^2}\\ c=\pm \sqrt{10+6}{\text{ [substitute a}}^2{\text{ by 10 and b}}^2\text{ by 6}]\\ c=\pm \sqrt{16}\text{ [square and add]}\\ c=\pm 4\text{ [square root]}\end{array}$

Foci: $c\left(0,\pm 4\right)$

  $\frac{y^2}{k}-\frac{x^2}{16-k}=1$

  $\begin{array}{l}\frac{y^2}{11}-\frac{x^2}{16-11}=1\text{ [substitute k=11]}\\ \frac{y^2}{11}-\frac{x^2}{5}=1\end{array}$

Foci :

  $\begin{array}{l}c=\pm \sqrt{a^2+b^2}\\ c=\pm \sqrt{11+5}{\text{ [substitute a}}^2{\text{ by 11 and b}}^2\text{ by 5}]\\ c=\pm \sqrt{16}\text{ [square and add]}\\ c=\pm 4\text{ [square root]}\end{array}$

Foci: $c\left(0,\pm 4\right)$

  $\frac{y^2}{k}-\frac{x^2}{16-k}=1$

  $\begin{array}{l}\frac{y^2}{12}-\frac{x^2}{16-12}=1\text{ [substitute k=12]}\\ \frac{y^2}{12}-\frac{x^2}{4}=1\end{array}$

Foci :

  $\begin{array}{l}c=\pm \sqrt{a^2+b^2}\\ c=\pm \sqrt{12+4}{\text{ [substitute a}}^2{\text{ by 12 and b}}^2\text{ by 4}]\\ c=\pm \sqrt{16}\text{ [square and add]}\\ c=\pm 4\text{ [square root]}\end{array}$

Foci: $c\left(0,\pm 4\right)$

  $\frac{y^2}{k}-\frac{x^2}{16-k}=1$

  $\begin{array}{l}\frac{y^2}{13}-\frac{x^2}{16-13}=1\text{ [substitute k=13]}\\ \frac{y^2}{13}-\frac{x^2}{3}=1\end{array}$

Foci :

  $\begin{array}{l}c=\pm \sqrt{a^2+b^2}\\ c=\pm \sqrt{13+3}{\text{ [substitute a}}^2{\text{ by 13 and b}}^2\text{ by 3}]\\ c=\pm \sqrt{16}\text{ [square and add]}\\ c=\pm 4\text{ [square root]}\end{array}$

Foci: $c\left(0,\pm 4\right)$

  $\frac{y^2}{k}-\frac{x^2}{16-k}=1$

  $\begin{array}{l}\frac{y^2}{14}-\frac{x^2}{16-14}=1\text{ [substitute k=14]}\\ \frac{y^2}{14}-\frac{x^2}{2}=1\end{array}$

Foci :

  $\begin{array}{l}c=\pm \sqrt{a^2+b^2}\\ c=\pm \sqrt{14+2}{\text{ [substitute a}}^2{\text{ by 14 and b}}^2\text{ by 2}]\\ c=\pm \sqrt{16}\text{ [square and add]}\\ c=\pm 4\text{ [square root]}\end{array}$

Foci: $c\left(0,\pm 4\right)$

  $\frac{y^2}{k}-\frac{x^2}{16-k}=1$

  $\begin{array}{l}\frac{y^2}{15}-\frac{x^2}{16-15}=1\text{ [substitute k=15]}\\ \frac{y^2}{15}-\frac{x^2}{1}=1\end{array}$

Foci :

  $\begin{array}{l}c=\pm \sqrt{a^2+b^2}\\ c=\pm \sqrt{15+1}{\text{ [substitute a}}^2{\text{ by 15 and b}}^2\text{ by 1}]\\ c=\pm \sqrt{16}\text{ [square and add]}\\ c=\pm 4\text{ [square root]}\end{array}$

Foci: $c\left(0,\pm 4\right)$

Because all hyperbolas has same foci, so they all are confocal.

b.

To determine

To graph the hyperbola using graphing device.

Explanation of Solution

Given information: Use graphing device to graph the hyperbola $\frac{y^2}{k}-\frac{x^2}{16-k}=1$ .

Graph: to draw the graph using a graphing device, it is needed to solve equation for y.

  $\begin{array}{l}{y}^2-\frac{x^2}{15}=1\\ y^2=1+\frac{x^2}{15}\text{ [subtract }\frac{x^2}{15}]\\ y^2=\frac{15+x^2}{15}\text{ [simplify]}\\ y=\pm \sqrt{\frac{15+x^2}{15}}\text{ [sqaure root]}\end{array}$

  $\begin{array}{l}\frac{y^2}{4}-\frac{x^2}{12}=1\\ \frac{y^2}{4}=1+\frac{x^2}{12}\text{ [subtract }\frac{x^2}{12}]\\ y^2=4·\frac{12+x^2}{12}\text{ [simplify]}\\ y=\pm 2\sqrt{\frac{12+x^2}{12}}\text{ [sqaure root]}\end{array}$

  $\begin{array}{l}\frac{y^2}{8}-\frac{x^2}{8}=1\\ \frac{y^2}{8}=1+\frac{x^2}{8}\text{ [subtract }\frac{x^2}{8}]\\ y^2=8·\frac{8+x^2}{8}\text{ [simplify]}\\ y=\pm \sqrt{8+x^2}\text{ [sqaure root]}\end{array}$

  $\begin{array}{l}\frac{y^2}{12}-\frac{x^2}{4}=1\\ \frac{y^2}{12}=1+\frac{x^2}{4}\text{ [subtract }\frac{x^2}{4}]\\ y^2=8·\frac{4+x^2}{4}\text{ [simplify]}\\ y=\pm \sqrt{8+2x^2}\text{ [sqaure root]}\end{array}$

To obtain the graph of the hyperbola, graph the following functions,

  $y_1=\pm \sqrt{\frac{15+x^2}{15}},y_2=\pm 2\sqrt{\frac{12+x^2}{12}},y_3=\pm \sqrt{8+x^2},y_4=\pm \sqrt{8+2x^2}$

Sketch the graph using graphing utility.

Step 1: Press WINDOW button to access the Window editor.

Step 2: Press $\boxed{\text{Y=}}$ button.

Step 3: Enter the expression $y_1=\pm \sqrt{\frac{15+x^2}{15}},y_2=\pm 2\sqrt{\frac{12+x^2}{12}},y_3=\pm \sqrt{8+x^2},y_4=\pm \sqrt{8+2x^2}$ which is required to graph.

Step 4: Press GRAPH button to graph the function.

The graph is obtained as:

  

Interpretation:

From the above graph it can be observed that the shape of the graph shrink as the value of k increases.