AP EAMCET · Maths · Hyperbola
If \(\theta\) is the angle subtended by a latus rectum at the centre of the hyperbola having eccentricity \(\frac{2}{\sqrt{7}-\sqrt{3}}\), then \(\sin \theta=\)
- A \(\frac{1}{2} \tan \frac{\theta}{2}\)
- B \(2 \cos \frac{\theta}{2}\)
- C \(\frac{1}{\sin \frac{\theta}{2}+\cos \frac{\theta}{2}}\)
- D \(1-\cos \frac{\theta}{2}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{2} \tan \frac{\theta}{2}\)
Step-by-step Solution
Detailed explanation
\(e = \frac{2}{\sqrt{7}-\sqrt{3}} \cdot \frac{\sqrt{7}+\sqrt{3}}{\sqrt{7}+\sqrt{3}} = \frac{2(\sqrt{7}+\sqrt{3})}{7-3} = \frac{\sqrt{7}+\sqrt{3}}{2}\) For a hyperbola, \(\tan(\frac{\theta}{2}) = \frac{b^2}{a^2e}\) Since \(b^2 = a^2(e^2-1)\),…
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