Hyperbola
Geometry of Hyperbola
JEE Advanced 2018 Paper 2
Grade 11
Question:
For the hyperbola $H : \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ where $a > b > 0$, whose conjugate axis $LM$ subtends an angle of $60^\circ$ at vertex $N$, and area of $\triangle LMN = 4\sqrt{3}$, find the length of the conjugate axis.
Step-by-Step Solution
Key Concept: Vertex $N = (a, 0); L = (0, b), M = (0, -b)$. $\angle LNM = 60^\circ \Rightarrow \tan 30^\circ = b/a \Rightarrow b = a/\sqrt{3}$. Area $= \frac{1}{2} \cdot 2b \cdot a = ab = 4\sqrt{3}$. Then $a \cdot a/\sqrt{3} = 4\sqrt{3} \Rightarrow a^2 = 12 \Rightarrow a = 2\sqrt{3}, b = 2$. Conjugate axis $= 2b = 4$.
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Correct Answer: 4