Question

The sides of a rectangle of greatest area which can be inscribed in an ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$

• A. 5, 3
• B. $2,\sqrt{2}$
• C. $5\sqrt{2},3\sqrt{2}$
• D. $3,\sqrt{2}$

Answer 1

Answer: (C)

$5\sqrt{2},3\sqrt{2}$

Let vertex be $(5sin\theta ,3cos\theta )(5sin\theta ,-3cos\theta )$
$(-5sin\theta ,3cos\theta )(-5sin\theta ,-3cos\theta )$
area of rectangle $=10sin\theta \times 6cos\theta$
$=15si{n}^2\theta$
area is max when $\theta =\frac{\pi }{4}$
So side $=10sin\theta$ & $6cos\theta$
$=5\sqrt{2},3\sqrt{2}$