InterviewSolution
Saved Bookmarks
| 1. |
The eccentricity of the hyperbola \(\rm \frac{x^2}{100} - \frac{y^2}{75} = 1\) is1. \(\sqrt { \frac{3}{4}}\)2. \(\sqrt { \frac{5}{4}}\)3. \(\sqrt { \frac{7}{4}}\)4. \(\sqrt { \frac{7}{3}}\) |
|
Answer» Correct Answer - Option 3 : \(\sqrt { \frac{7}{4}}\) Concept: Standard equation of an hyperbola : \(\frac{{{\rm{\;}}{{\bf{x}}^2}}}{{{{\bf{a}}^2}}} - \frac{{{{\bf{y}}^2}}}{{{{\bf{b}}^2}}} = 1\)
Calculation: Given: \(\rm \frac{x^2}{100} - \frac{y^2}{75} = 1\) 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 = 100 and b2 = 75 Now, Eccentricity (e) = \(\sqrt {1 + {\rm{\;}}\frac{{{{\rm{b}}^2}}}{{{{\rm{a}}^2}}}} \) = \(\sqrt {1 + \frac{75}{100}}\) = \(\sqrt {1 + \frac{3}{4}}\) = \(\sqrt { \frac{7}{4}}\) |
|