Hyperbola
For the hyperbola written as x^2/9 - y^2/16 = 1, find its eccentricity and the equations of the two asymptotes that bound the curve.
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About This Question
- Subject
- mathematics
- Chapter
- coordinate geometry
- Topic
- hyperbola
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
e=5/3,asymptotesy=±(4/3)x
For a hyperbola x^2/a^2 - y^2/b^2 = 1, the eccentricity obeys b^2 = a^2(e^2 - 1) and the asymptotes are y = ±(b/a)x, so reading a^2 and b^2 settles both. Here a^2 = 9 and b^2 = 16, giving a = 3 and b = 4. Then e^2 = 1 + b^2/a^2 = 1 + 16/9 = 25/9, so e = 5/3. The asymptotes are y = ±(b/a)x = ±(4/3)x. Option e = 4/3 mistakenly uses b/a directly as the eccentricity. Option e = 5/4 swaps a and b in the ratio. Option asymptotes y = ±(3/4)x inverts the slope b/a to a/b. This applies the standard JEE Advanced hyperbola formulas. Plausibility check: a hyperbola must satisfy e > 1, and 5/3 ≈ 1.67 qualifies; moreover c = ae = 5 satisfies c^2 = a^2 + b^2 = 9 + 16 = 25, confirming the focal and asymptotic data agree.
This medium difficulty mathematics question is from the chapter coordinate geometry, covering the topic of hyperbola. It appeared in the 2025 exam.
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