AP EAMCET · Maths · Ellipse
Let \(\mathrm{A}, \mathrm{A}^{\prime}\) be the end points of major axis, \(\mathrm{S}, \mathrm{S}^{\prime}\) be the foci and \(\mathrm{B}, \mathrm{B}^{\prime}\) be the end points of minor axis of an ellipse \(\mathrm{E}\). If \(\angle \mathrm{BAB}^{\prime}=60^{\circ}\), then \(\angle \mathrm{SBS}^{\prime}=\)
- A \(\tan ^{-1}(\sqrt{2})\)
- B \(\tan ^{-1}(-2 \sqrt{2})\)
- C \(\tan ^{-1}\left(\sqrt{\frac{2}{3}}\right)\)
- D \(\tan ^{-1}\left(\sqrt{\frac{3}{2}}\right)\)
Answer & Solution
Correct Answer
(B) \(\tan ^{-1}(-2 \sqrt{2})\)
Step-by-step Solution
Detailed explanation
\(\tan 30^{\circ}=\frac{b}{a} \Rightarrow a=\sqrt{3} b\) We have \(c=\sqrt{a^2-b^2}= \pm \sqrt{2} b\) Now, \(\tan \theta=\frac{c}{b}=\frac{ \pm \sqrt{2} b}{b}= \pm \sqrt{2}\),…
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