Question

The sides of the rectangle of the greatest area, that can be inscribed in the ellipse x² + 2y² = 8, are given by –

• A. $4 \sqrt { 2 }$ , 4
• B. 4, $2 \sqrt { 2 }$
• C. 2, $\sqrt { 2 }$
• D. ) 2 $\sqrt { 2 }$ , 2

Answer 1

Answer: (B)

4, $2 \sqrt { 2 }$

Answer 2

Any point on the ellipse + = 1 is (2 $\sqrt { 2 }$ cos q, 2 sin q). (See Fig.)

A = Area of the inscribed rectangle

= 4(2$\sqrt { 2 }$ cos q) (2 sin q)

= 8$\sqrt { 2 }$ sin 2q

$\frac { \mathrm { dA } } { \mathrm { d } \theta }$ = 16$\sqrt { 2 }$cos 2q = 0 Ž q = p/4

Also = –32$\sqrt { 2 }$sin 2q < 0 for q = p/4.