AP EAMCET · Maths · Ellipse
The sides of the rectangle of greatest area that can be inscribed in the ellipse \(x^2+4 y^2=64\) are :
- A \((6 \sqrt{2}, 4 \sqrt{2})\)
- B \((8 \sqrt{2}, 4 \sqrt{2})\)
- C \((8 \sqrt{2}, 8 \sqrt{2})\)
- D \((16 \sqrt{2}, 4 \sqrt{2})\)
Answer & Solution
Correct Answer
(B) \((8 \sqrt{2}, 4 \sqrt{2})\)
Step-by-step Solution
Detailed explanation
Given ellipse: \(x^2+4 y^2=64 \implies \frac{x^2}{64} + \frac{y^2}{16} = 1\) \(a^2=64 \implies a=8\) \(b^2=16 \implies b=4\) For greatest area, half-sides are \(x=\frac{a}{\sqrt{2}}\) and \(y=\frac{b}{\sqrt{2}}\). Side 1 \(= 2x = a\sqrt{2} = 8\sqrt{2}\) Side 2…
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