Question

$Thefociofthehyperbola2x^2-3y^2=5are$

• A. $(\pm 5/\sqrt{6}.0)$
• B. $(\pm 5/6.0)$
• C. $(\pm \sqrt{5}/6,0)$
• D. none of these

Answer 1

The correct option is A

$(\pm 5/\sqrt{6}.0)$

$\text{the given equation of hyperbola is}2x^2-3y^2=5.\text{It can be rewritten in the following :}$

$\frac{2x^2}{5}-\frac{3y^2}{5}=1$

$\Rightarrow \frac{x^2}{\frac{5}{2}}-\frac{y^2}{\frac{5}{2}}=1$

$\text{This is the standard equation of a paralbola,where}{a}^2=\frac{5}{2}and{b}^2=\frac{5}{3}$

$\text{The eccentricity can be calculated in the following ways :}$

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

$\Rightarrow \frac{5}{3}=\frac{5}{2}(e^2-1)$

$\Rightarrow e^2-1=\frac{2}{3}$

$\Rightarrow e^2=\frac{5}{3}$

$\Rightarrow e=\sqrt{\frac{5}{3}}$

$Coordinatesofthefoci=(\pm ae,0)=(\pm 5/\sqrt{6},0)$