Chapter A5.2, Problem 15E

To determine

To find: the foci, eccentricity and directrix of ellipse

Answer to Problem 15E

Eccentricity is $e=\sqrt{\frac{5}{4}}$

Foci is $[F=(\pm 5,0)]$

Directrixis $x=0\pm \frac{16}{5}$

Explanation of Solution

Given:

  $9x^2-16y^2=144,$

Calculation:

Given with the equation of hyperbola as

  $9x^2-16y^2=144,$

which could also be written as $\frac{x^2}{16}-\frac{y^2}{9}=1$

The parameters of hyperbola can be calculated as:

  $c=\sqrt{a^2+b^2}=\sqrt{16+9}=5$

Eccentricity $e=\frac{c}{a}$ therefore,

  $e=\sqrt{\frac{5}{4}}$

Asymptotes of the hyperbola can be calculated as $y=\pm \frac{3}{4}x$

Focus $[F=(\pm 5,0)]$

Directrixis $x=0\pm \frac{16}{5}$

Conclusion:

Eccentricity is $e=\sqrt{\frac{5}{4}}$

Foci is $[F=(\pm 5,0)]$

Directrixis $x=0\pm \frac{16}{5}$