TS EAMCET · Maths · Ellipse
An ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b)\) is inscribed in a rectangle of dimensions \(2 a\) and \(2 b\) respectively, If the angle between the diagonals of the rectangle is \(\tan ^{-1}(4 \sqrt{3})\), then the eccentricity of that ellipse is
- A \(\frac{1}{\sqrt{2}}\)
- B \(\frac{1}{2}\)
- C \(\frac{1}{\sqrt{3}}\)
- D \(\frac{\sqrt{2}}{\sqrt{3}}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
Now, from the figure \(\tan \theta=\frac{b}{a}\) \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) Since, \(\tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^2 \theta}\) [where \(2 \theta\) is angle between the diagonals of the rectangle]…
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