Question:
The eccentricity of the hyperbola \(16x^2 – 9y^2 = 1\) is:
अतिपरवलय \(16x^2 – 9y^2 = 1\) की उत्केंद्रता (eccentricity) है:
Options:
- 3/5
- 5/3
- 4/5
- 5/4
- 3/5
- 5/3
- 4/5
- 5/4
Correct Answer: b)
Explanation:
Writing the hyperbola in standard form \(\frac{x^2}{a^2} – \frac{y^2}{b^2} = 1\):
\[\frac{x^2}{(1/16)} – \frac{y^2}{(1/9)} = 1 \implies a^2 = \frac{1}{16}, \; b^2 = \frac{1}{9}\]
The eccentricity \(e\) of a hyperbola is given by:
\[e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{1/9}{1/16}} = \sqrt{1 + \frac{16}{9}} = \sqrt{\frac{25}{9}} = \frac{5}{3}\]
Hence, the correct option is (b).
अतिपरवलय को मानक रूप \(\frac{x^2}{a^2} – \frac{y^2}{b^2} = 1\) में लिखने पर:
\[a^2 = \frac{1}{16}, \quad b^2 = \frac{1}{9}\]
उत्केंद्रता \(e\) का सूत्र:
\[e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{1/9}{1/16}} = \sqrt{1 + \frac{16}{9}} = \sqrt{\frac{25}{9}} = \frac{5}{3}\]
अतः सही विकल्प (b) है।