Question
Download Solution PDFThe eccentricity of the hyperbola, 16x2 – 9y2 = 1.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFI. The equation of a hyperbola with the centre at origin and foci on x – axis is given by:
\(\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1\) and the eccentricity \(e = \sqrt {1 + \frac{{{b^2}}}{{{a^2}}}}\)
II. The equation of a hyperbola with the centre at origin and foci on y – axis is given by:
\(\frac{{{y^2}}}{{{b^2}}} - \frac{{{x^2}}}{{{a^2}}} = 1\) and the eccentricity \(e = \sqrt {1 + \frac{{{a^2}}}{{{b^2}}}}\)
Calculation:
Given: Equation of hyperbola is 16x2 – 9y2 = 1
The given equation of hyperbola can be written as: \(\frac{{{x^2}}}{{\frac{1}{{16}}}} - \frac{{{y^2}}}{{\frac{1}{9}}} = 1\)
Here, a2 = 1/16 and b2 = 1/9.
As we know that, the equation of a hyperbola with the centre at origin and foci on x – axis is given by:
\(\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1\) and the eccentricity \(e = \sqrt {1 + \frac{{{b^2}}}{{{a^2}}}}\)
\(\Rightarrow e = \sqrt {1 + \frac{{\frac{1}{9}}}{{\frac{1}{{16}}}}} = \frac{5}{3}\)Last updated on Jul 11, 2025
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