AP EAMCET · Maths · Hyperbola
Let \(L L^{\prime}\) be the latus rectum through the focus \(S\) of a hyperbola and \(A^{\prime}\) be the opposite vertex of the hyperbola. If triangle \(A^{\prime} L L^{\prime}\) is equilateral, then the eccentricity of the hyperbola is
- A \(\frac{\sqrt{3}+1}{\sqrt{3}}\)
- B \(\frac{\sqrt{3}+1}{\sqrt{2}}\)
- C \(\frac{\sqrt{3}+1}{\sqrt{5}}\)
- D \(\sqrt{3}+1\)
Answer & Solution
Correct Answer
(A) \(\frac{\sqrt{3}+1}{\sqrt{3}}\)
Step-by-step Solution
Detailed explanation
Length of latus rectum \(LL' = \frac{2b^2}{a}\). Height of triangle \(A'LL'\) from \(A'(-a,0)\) to the line \(x=ae\) is \(h = ae - (-a) = a(e+1)\). For an equilateral triangle, \(h = \frac{\sqrt{3}}{2} \cdot LL'\). \(a(e+1) = \frac{\sqrt{3}}{2} \cdot \frac{2b^2}{a}\)…
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