Question
Download Solution PDFThe eccentricity of the hyperbola 16x2 – 9y2 = 1 is
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Hyperbola: The locus of a point which moves such that its distance from a fixed point is greater than its distance from a fixed straight line. (Eccentricity = e > 1)
Equation |
\(\frac{{\;{{\bf{x}}^2}}}{{{{\bf{a}}^2}}} - \frac{{{{\bf{y}}^2}}}{{{{\bf{b}}^2}}} = 1\) |
\(- \frac{{\;{{\bf{x}}^2}}}{{{{\bf{a}}^2}}} + \frac{{{{\bf{y}}^2}}}{{{{\bf{b}}^2}}} = 1\) |
Equation of Transverse axis |
y = 0 |
x = 0 |
Equation of Conjugate axis |
x = 0 |
y = 0 |
Length of Transverse axis |
2a |
2b |
Length of Conjugate axis |
2b |
2a |
Vertices |
(± a, 0) |
(0, ± b) |
Focus |
(± ae, 0) |
(0, ± be) |
Directrix |
x = ± a/e |
y = ± b/e |
Centre |
(0, 0) |
(0, 0) |
Eccentricity |
\(\sqrt {1 + {\rm{\;}}\frac{{{{\rm{b}}^2}}}{{{{\rm{a}}^2}}}}\) |
\(\sqrt {1 + {\rm{\;}}\frac{{{{\rm{a}}^2}}}{{{{\rm{b}}^2}}}}\) |
Length of Latus rectum |
\(\frac{{2{{\rm{b}}^2}}}{{\rm{a}}}\) |
\(\frac{{2{{\rm{a}}^2}}}{{\rm{b}}}\) |
Focal distance of the point (x, y) |
ex ± a |
ey ± a |
Calculation:
Given:
16x2 – 9y2 = 1
\( \Rightarrow \frac{{{\rm{\;}}{{\rm{x}}^2}}}{{\frac{1}{{16}}}} - \frac{{{{\rm{y}}^2}}}{{\frac{1}{9}}} = 1\)
Compare with \(\frac{{{\rm{\;}}{{\rm{x}}^2}}}{{{{\rm{a}}^2}}} - \frac{{{{\rm{y}}^2}}}{{{{\rm{b}}^2}}} = 1\)
∴ a2 = 1/16 and b2 = 1/9
Eccentricity = \(\sqrt {1 + {\rm{\;}}\frac{{{{\rm{b}}^2}}}{{{{\rm{a}}^2}}}} = \;\sqrt {1 + \;\frac{{\left( {\frac{1}{9}} \right)}}{{\left( {\frac{1}{{16}}} \right)}}} = \;\sqrt {1 + \;\frac{{16}}{9}} = \;\sqrt {\frac{{25}}{9}} = \;\frac{5}{3}\)
Last updated on May 30, 2025
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