WBJEE · Maths · Ellipse
Area in the first quadrant between the ellipses \(x^{2}+2 y^{2}=a^{2}\) and \(2 x^{2}+y^{2}=a^{2}\) is
- A \(\frac{a^{2}}{\sqrt{2}} \tan ^{-1} \frac{1}{\sqrt{2}}\)
- B \(\frac{3 a^{2}}{4} \tan ^{-1} \frac{1}{2}\)
- C \(\frac{5 a^{2}}{2} \sin ^{-1} \frac{1}{2}\)
- D \(\frac{9 \pi a^{2}}{2}\)
Answer & Solution
Correct Answer
(A) \(\frac{a^{2}}{\sqrt{2}} \tan ^{-1} \frac{1}{\sqrt{2}}\)
Step-by-step Solution
Detailed explanation
Hint: \(A=\left(\frac{a^{2}}{6}+\int_{a / \sqrt{3}}^{a / \sqrt{2}} \sqrt{a^{2}-2 x^{2}} d x\right) \times 2=\frac{a^{2}}{\sqrt{2}} \tan ^{-1}\left(\frac{1}{\sqrt{2}}\right)\)
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