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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 :

  1. A \((6 \sqrt{2}, 4 \sqrt{2})\)
  2. B \((8 \sqrt{2}, 4 \sqrt{2})\)
  3. C \((8 \sqrt{2}, 8 \sqrt{2})\)
  4. D \((16 \sqrt{2}, 4 \sqrt{2})\)
Verified Solution

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…