Chapter 9.5, Problem 45HP

To determine

To Find: The equation of the hyperbola.

Answer to Problem 45HP

  $\frac{{\left(x-3\right)}^2}{5}-\frac{{\left(y+2\right)}^2}{5}=1$

Explanation of Solution

Given:

  $\begin{array}{l}\text{Origin lies on a horizontal hyperbola}\text{.}\\ y=\left(-x+1\right)\&\left(x-5\right)\end{array}$

Formula Used:

  $\begin{array}{l}\frac{{\left(y-k\right)}^2}{a^2}-\frac{{\left(x-h\right)}^2}{b^2}=1\\ \frac{y^2}{a^2}-\frac{x^2}{b^2}=1\\ y=\pm \frac{b}{a}\left(x-h\right)+k\\ e=\frac{\sqrt{a^2+b^2}}{a}\\ l=\frac{2b^2}{a}\\ Vertices:\left(h-a,k\right)\&\left(h+a,k\right)\\ Foci:\left(h,k-c\right)\&\left(h,k+c\right)\end{array}$

Calculation:

  $\begin{array}{l}\text{Origin lies on a horizontal hyperbola}\text{.}\\ y=\left(-x+1\right)\&\left(x-5\right)\\ \frac{{\left(y-k\right)}^2}{a^2}-\frac{{\left(x-h\right)}^2}{b^2}=1\\ y=\pm \frac{b}{a}\left(x-h\right)+k\\ \frac{b}{a}=1\Rightarrow a=b\\ \left(-\frac{b}{a}\right)\left(-h\right)+k=1\\ \left(\frac{b}{a}\right)\left(-h\right)+k=-5\\ h+k=1\\ -h+k=-5\\ 2k=-4\\ \Rightarrow k=-2\\ \Rightarrow h=3\\ \frac{{\left(0-3\right)}^2}{a^2}-\frac{{\left(0+2\right)}^2}{a^2}=1\\ \Rightarrow a^2=5\\ \frac{{\left(x-3\right)}^2}{5}-\frac{{\left(y+2\right)}^2}{5}=1\end{array}$

Conclusion:

Thus the equation is $\frac{{\left(x-3\right)}^2}{5}-\frac{{\left(y+2\right)}^2}{5}=1$ .