Question

The eccentricity of the hyperbola \(\rm \frac{x^2}{100} - \frac{y^2}{75} = 1\) is
1. \(\sqrt { \frac{3}{4}}\)
2. \(\sqrt { \frac{5}{4}}\)
3. \(\sqrt { \frac{7}{4}}\)
4. \(\sqrt { \frac{7}{3}}\)

Answer 1

Correct Answer - Option 3 : \(\sqrt { \frac{7}{4}}\)

Concept:

Standard equation of an hyperbola : \(\frac{{{\rm{\;}}{{\bf{x}}^2}}}{{{{\bf{a}}^2}}} - \frac{{{{\bf{y}}^2}}}{{{{\bf{b}}^2}}} = 1\) 

• 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: \(\rm \frac{x^2}{100} - \frac{y^2}{75} = 1\)

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) = \(\sqrt {1 + {\rm{\;}}\frac{{{{\rm{b}}^2}}}{{{{\rm{a}}^2}}}} \) 

= \(\sqrt {1 + \frac{75}{100}}\)

= \(\sqrt {1 + \frac{3}{4}}\)

= \(\sqrt { \frac{7}{4}}\)