Correct Answer - Option 1 : (± 5, 0)
Concept:
Standard equation of an hyperbola: \(\frac{{{\rm{\;}}{{\bf{x}}^2}}}{{{{\bf{a}}^2}}} - \frac{{{{\bf{y}}^2}}}{{{{\bf{b}}^2}}} = 1\) (a > b)
- Coordinates of foci = (± ae, 0)
- Eccentricity (e) = \(\sqrt {1 + {\rm{\;}}\frac{{{{\rm{b}}^2}}}{{{{\rm{a}}^2}}}} \) ⇔ a2e2 = a2 + b2
- Length of Latus rectum = \(\rm \frac{2b^2}{a}\)
Calculation:
Given: \(\rm \frac{x^2}{16} - \frac{y^2}{9} = 1\)
Compare with the standard equation of a hyperbola: \(\frac{{{\rm{\;}}{{\bf{x}}^2}}}{{{{\bf{a}}^2}}} - \frac{{{{\bf{y}}^2}}}{{{{\bf{b}}^2}}} = 1\)
So, a2 = 16 and b2 = 9
⇒ a = 4 and b = 3 (a > b)
Now, Eccentricity (e) = \(\sqrt {1 + {\rm{\;}}\frac{{{{\rm{b}}^2}}}{{{{\rm{a}}^2}}}} \)
= \(\sqrt {1 + \frac{9}{16}}\)
= \(\sqrt { \frac{16+9}{16}}\)
= \(\frac 54\)
Coordinates of foci = (± ae, 0)
= (± 5, 0)
