Question

find the eccentricity of the hyperbola 16x^2-9y^2=144​

Answer 1

$\frac{16}{144} x - \frac{9}{144} y = \frac{144}{144}$

$\frac{1}{9} x - \frac{1}{16} y = 1$

a^2=9 b^2=16

a=3 b=4

eccentricity e=

$\sqrt { {a}^{2} + {b}^{2} } \div {b}^{2}$

$\sqrt{9 + 16 \div 16 } = \sqrt{25 \div 16} = \frac{5}{4}$