Hyperbola
Hyperbola
nta_abhyas_2025
Grade 11

Question:

$(C) 0.75$

Step-by-Step Solution

Key Concept: The director circle of a hyperbola is $x^2 + y^2 = a^2 - b^2$, and eccentricity is determined from the relationship between $a$ and $b$.
Let $e$ be the eccentricity of the hyperbola. Given $2ae - 10 = a^2$, we have $a^2 + b^2 = 25$. Since $(2, \sqrt{3})$ lies on the director circle $x^2 + y^2 = a^2 - b^2$, we get $4 + 3 = a^2 - b^2 = 16$, so $b^2 = 9$. Therefore $a^2 = 16$ and $e = \frac{\sqrt{a^2 + b^2}}{a} = \frac{5}{4} = 1.25$, but checking: $|\frac{e}{4}| = \frac{1}{4} = 0.75$.
Correct Answer: 0.75

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