Question

Find the equation of the hyperbola whose foci are $(2,3)$ and $(-4,3)$ and eccentricity is $5/4$

Answer 1

Let equation be$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Foci are $(2,3)$ and $(-4,3)$ and $e=\frac{5}{4}$

the distance between foci of hyperbola$=2ae$

$\therefore 2ae=\sqrt{(2+4{)}^2+(3-3{)}^2}$

$2ae=6\Rightarrow 2\times a\times \frac{5}{4}=6$

$a=\frac{12}{5}$

$b^2=a^2(e^2-1)$

$b^2={\frac{12}{5}}^2({\frac{5}{4}}^2-1)$

$b^2=\frac{144}{25}\left(\frac{9}{16}\right)$

$b^2=\frac{9\times 9}{25}$

$b=\frac{9}{5}$

Thus, equation of hyperbola is$\frac{x^2}{{\left(\frac{12}{5}\right)}^2}-\frac{y^2}{{\left(\frac{9}{5}\right)}^2}=1$