Chapter 10.6, Problem 66E

To determine

To find: a set of parametric equation to represent the section of the graph of the line or conic.

Answer to Problem 66E

The positive sign will be the equation of right branch of the hyperbola $y=3+\pm 12\sqrt{\frac{{\left(x-1\right)}^2}{25}-1}$ .

Explanation of Solution

Given information:

Right branch of the hyperbola with vertices $\left(-4,3\right)$ and $\left(6,3\right)$ and foci $\left(-12,3\right)$ and $\left(14,3\right)$

Calculation:

The vertices of hyperbola are $\left(-4,3\right)$

  $\left(6,3\right)$ and the foci of hyperbola is $\left(-12,3\right)$

  $\left(14,3\right)$ .

Since, the $x$ coordinates are changing for vertices and foci so, the hyperbola is a horizontal hyperbola. The center of the hyperbola will be the midnight of the vertices so,

  $\left(h,k\right)=\left(\frac{-4+6}{2},\frac{3+3}{2}\right)\Rightarrow \left(h,k\right)=\left(1,3\right)$ . Therefore, $a=6-1\Rightarrow a^2=25$

Since, the foci are at $\left(-12,3\right)$ and $\left(14,3\right)$ so, the center of foci will be

  $\left(\frac{-12+14}{2},\frac{3+3}{2}\right)=\left(1,3\right)$ . Therefore, $c=14-1\Rightarrow c^2=169$

Use $c^2=a^2+b^2$ to find $b^2$ .

  $\begin{array}{l}{c}^2=a^2+b^2\\ 169=25+b^2\\ b^2=144\end{array}$

Substitute the values into the equation of the hyperbola $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$

  $\begin{array}{c}\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1\\ \frac{{\left(x-1\right)}^2}{25}-\frac{{\left(y-3\right)}^2}{144}=1\\ \frac{{\left(y-3\right)}^2}{144}=\frac{{\left(x-1\right)}^2}{25}-1\\ y-3=\pm 12\sqrt{\frac{{\left(x-1\right)}^2}{25}-1}\\ y=3+\pm 12\sqrt{\frac{{\left(x-1\right)}^2}{25}-1}\end{array}$

Thus, the positive sign will be the equation of right branch of the hyperbola $y=3+\pm 12\sqrt{\frac{{\left(x-1\right)}^2}{25}-1}$ .