AP EAMCET · Maths · Ellipse
The lengths of 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
Equation of given ellipse is \(\begin{array}{rlrl} & x^2+4 y^2 =64 \\ \Rightarrow & \frac{x^2}{64}+\frac{y^2}{16} =1 \end{array}\) Let the vertex \(P(8 \cos \theta, 4 \sin \theta)\) of the rectangle \(P Q R S\) having greatest area.…
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