AP EAMCET · Maths · Hyperbola
One of the latus recta of the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) subtends an angle \(2 \operatorname{Tan}^{-1}\left(\frac{3}{2}\right)\) at the centre of the hyperbola. If \(\mathrm{b}^2=36\) and e is the eccentricity of the given hyperbola, then \(\sqrt{\mathrm{a}^2+\mathrm{e}^2}=\)
- A \(4\)
- B \(\sqrt{14}\)
- C \(6\)
- D \(\sqrt{21}\)
Answer & Solution
Correct Answer
(A) \(4\)
Step-by-step Solution
Detailed explanation
\(\frac{b^2}{a^2e} = \frac{3}{2}\) \(\frac{36}{a^2e} = \frac{3}{2} \Rightarrow a^2e = 24\) \(b^2 = a^2(e^2-1) \Rightarrow 36 = a^2e^2 - a^2\) \(36 = (a^2e)e - a^2 \Rightarrow 36 = 24e - a^2\) \(a^2 = 24e - 36\) Substitute \(a^2\) into \(a^2e=24\): \((24e-36)e = 24\)…
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