Chapter 9.5, Problem 14WE

To determine

To calculate: To find the equation of hyperbola

Answer to Problem 14WE

Equation of hyperbola is $\frac{x^2}{16}-\frac{y^2}{12}=1$

Explanation of Solution

Given information: Foci are $\left(-4,0\right)$ and $\left(4,0\right)$ and difference of focal radii is $4$

Formula Used:

The hyperbola with center $\left(0,0\right)$ and foci $\left(-c,0\right)$ and $\left(c,0\right)$ and difference of focal radii $2a$ has the equation:

  $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ , where $b^2=c^2-a^2$

Equation of the asymptotes are $y=\frac{b}{a}x$ and $y=-\frac{b}{a}x$

The hyperbola with center $\left(0,0\right)$ and foci $\left(0,-c\right)$ and $\left(0,c\right)$ and difference of focal radii $2a$ has the equation:

  $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ , where $b^2=c^2-a^2$

Equation of the asymptotes are $y=\frac{a}{b}x$ and $y=-\frac{a}{b}x$

Calculation:

Given that Foci of hyperbola are

  $\left(-4,0\right)$ and $\left(4,0\right)$

Thus,

  $c=\pm 4$

Also, given that difference of focal radii is $10$

Thus,

  $\begin{array}{l}2a=4\\ a=\frac{4}{2}\\ a=2\end{array}$

  $b$ is calculated as follows:

  $b^2=c^2-a^2$

Substituting the values:

  $\begin{array}{l}{b}^2={\left(\pm 4\right)}^2-{\left(2\right)}^2\\ b^2=16-4\\ b^2=12\\ b=\pm 2\sqrt{3}\end{array}$

Standard form of the equation of hyperbola is given as follows:

  $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Substituting the value:

  $\begin{array}{l}\frac{x^2}{4^2}-\frac{y^2}{{\left(2\sqrt{3}\right)}^2}=1\\ \frac{x^2}{16}-\frac{y^2}{12}=1\end{array}$

Conclusion:

Hence, equation of hyperbola is $\frac{x^2}{16}-\frac{y^2}{12}=1$