Question

Equation of the hyperbola whose vertices are $(\pm 3,0)$ and foci at $(\pm 5,0)$ is

• A. $16x^2-9y^2=144$
• B. $9x^2-16y^2=144$
• C. $25x^2-9y^2=225$
• D. $9x^2-25y^2=81$

Answer 1

The correct option is A

$16x^2-9y^2=144$

$\text{The vertices of the hyperbola are}(\pm 3,0)andfociare(\pm 5,0)$

$\text{Thus,the values of a and ae are 3 and 5,respectively.}$

$Now,usingtherelation{b}^2=a^2(e^2-1).weget$

$b^2=25-9$

$\Rightarrow b^2=16$

$\text{Equation of the hyperbola is given below:}$

$\frac{x^2}{9}-\frac{y^2}{16}=1$

$\Rightarrow 16x^{2}2-9y^2=144$