Question

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

Options

• 16x² − 9y² = 144

• 9x² − 16y² = 144

•  25x² − 9y² = 225

• 9x² − 25y² = 81

Answer 1

 16x² − 9y² = 144

The vertices of the hyperbola are  \[\left( \pm 3, 0 \right)\]  and foci are  \[\left( \pm 5, 0 \right)\].

Thus, the values of a and ae are 3 and 5, respectively. 
Now, using the relation 

\[b^2 = a^2 ( e^2 - 1)\], we get:

\[b^2 = 25 - 9\]

\[ \Rightarrow b^2 = 16\]

Equation of the hyperbola is given below: 

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

\[ \Rightarrow 16 x^2 - 9 y^2 = 144\]