Question

The foci of the hyperbola 9x² − 16y² = 144 are

विकल्प

• (± 4, 0)

• (0, ± 4)

•  (± 5, 0)

• (0, ± 5)

Answer 1

(± 5, 0)

The equation of the hyperbola is given below:

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

This equation can be rewritten in the following way:

\[\frac{9 x^2}{144} - \frac{16 y^2}{144} = 1\]

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

This is the standard equation of a hyperbola, where 

\[a^2 = 16 \text { and } b^2 = 9\]

The eccentricity is calculated in the following way:

\[b^2 = a^2 ( e^2 - 1)\]

\[ \Rightarrow 9 = 16( e^2 - 1)\]

\[ \Rightarrow \frac{9}{16} = e^2 - 1\]

\[ \Rightarrow e = \frac{5}{4}\]

Foci = \[\left( \pm ae, 0 \right) = \left( \pm 5, 0 \right)\]