Question Number 76811 by Rio Michael last updated on 30/Dec/19
$$\mathrm{The}\:\mathrm{eccentricity}\:\mathrm{of}\:\mathrm{the}\:\mathrm{hyperbola} \\ $$$$\:\:\:\frac{{x}^{\mathrm{2}} }{\mathrm{64}}\:−\:\frac{{y}^{\mathrm{2}} }{\mathrm{36}}\:=\:\mathrm{1}\:\mathrm{is}\: \\ $$$$\mathrm{A}.\:\frac{\mathrm{5}}{\mathrm{4}} \\ $$$$\mathrm{B}.\:\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\mathrm{C}.\:\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\mathrm{D}.\:\frac{\mathrm{4}}{\mathrm{3}} \\ $$
Commented by MJS last updated on 31/Dec/19
![e^2 =a^2 +b^2 a=8; b=6 ⇒ e=10 [this is the linear eccentricity] ε=(e/a)=(5/4) [this is the numeric eccentricity]](https://www.tinkutara.com/question/Q76838.png)
$${e}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} \\ $$$${a}=\mathrm{8};\:{b}=\mathrm{6} \\ $$$$\Rightarrow \\ $$$${e}=\mathrm{10}\:\left[\mathrm{this}\:\mathrm{is}\:\mathrm{the}\:\mathrm{linear}\:\mathrm{eccentricity}\right] \\ $$$$\epsilon=\frac{{e}}{{a}}=\frac{\mathrm{5}}{\mathrm{4}}\:\left[\mathrm{this}\:\mathrm{is}\:\mathrm{the}\:\mathrm{numeric}\:\mathrm{eccentricity}\right] \\ $$
Commented by Rio Michael last updated on 31/Dec/19
$${thanks}\:{sir} \\ $$