Chapter 9.5, Problem 43PPS

To determine

To Find: The equation for the horizontal hyperbola on which the lightning is located.

Answer to Problem 43PPS

  $\begin{array}{l}\Rightarrow \frac{{\left(x-4000\right)}^2}{2722500}-\frac{y^2}{13277500}=1\left(\text{equation 1}\right)\\ \Rightarrow \frac{{\left(y-2500\right)}^2}{1210000}-\frac{x^2}{5040000}=1\left(\text{equation 2}\right)\end{array}$

Explanation of Solution

Given:

  $\begin{array}{l}\text{Speed of sound}=1100feetper\sec ond\\ \text{meg hear }2\sec \text{ after lae and caser hears it }3\sec \text{ after lae}\end{array}$

  

Formula Used:

  $\begin{array}{l}{c}^2=a^2+b^2\\ \text{Hyperbola Standard equation: }\frac{x^2}{a^2}-\frac{y^{{}^2}}{b^2}=1\end{array}$

Calculation:

  $\to \text{In case of Caser \& Lae :}$

  $\text{on assuming foci at L \& C such that the hyperbola is horizontal \& it is centred at }\left(h,k\right)\to \left(4000,0\right)\text{: }$

  $\frac{{\left(x-4000\right)}^2}{a^2}-\frac{y^2}{b^2}=1$

  $\Rightarrow \text{Distance between them =}8000=2c$

  $\Rightarrow c=4000$

  $\to \text{Speed of sound }=1100feet/\sec ond$

  $\to \text{Time taken to hear the sound}=3\sec ond$

  $\to \text{Distance travelled by sound}=3\left(1100\right)=3300ft=\text{it is equal to fixed distance }2a$

  $\Rightarrow 2a=3300$

  $\Rightarrow a=1650$

  $\to a^2=2722500$

  $\to \text{using equation }{c}^2=a^2+b^2\text{ to solve for }{b}^2$

  $\Rightarrow b^2=c^2-a^2$

  $\Rightarrow b^2={4000}^2-{1650}^2=13277500$

  $\text{Now the equation is :}$

  $\Rightarrow \frac{{\left(x-4000\right)}^2}{2722500}-\frac{y^2}{13277500}=1\left(\text{equation 1}\right)$

  $\to \text{In case of Lae \& Meg :}$

  $\text{on assuming foci be at L \& M so the hyperbola is vertical and centered at }\left(h,k\right)\to \left(0,2500\right):$

  $\Rightarrow \frac{{\left(y-2500\right)}^2}{a^2}-\frac{x^2}{b^2}=1$

  $\to \text{Distance between them}=5000=2c$

  $\Rightarrow c=2500$

  $\to \text{Speed of sound }=1100feet/\sec ond$

  $\to \text{Time taken to hear the sound}=2\sec ond$

  $\to \text{Distance travelled by sound}=2\left(1100\right)=2200ft=\text{it is equal to fixed distance }2a:$

  $2a=2200\to a=1100$

  $\to a^2=1210000$

  $\to \text{using equation }{c}^2=a^2+b^2\text{ to solve for }{b}^2$

  $\Rightarrow b^2=c^2-a^2={2500}^2-{1100}^2=5040000$

  $\text{Equation is :}$

  $\Rightarrow \frac{{\left(y-2500\right)}^2}{1210000}-\frac{x^2}{5040000}=1\left(\text{equation 2}\right)$

Conclusion:

Thus the equation of the hyperbola is $\begin{array}{l}\Rightarrow \frac{{\left(x-4000\right)}^2}{2722500}-\frac{y^2}{13277500}=1\left(\text{equation 1}\right)\\ \Rightarrow \frac{{\left(y-2500\right)}^2}{1210000}-\frac{x^2}{5040000}=1\left(\text{equation 2}\right)\end{array}$