The greatest area of a rectangle, with sides parallel to the coordinate axes, that can be inscribed in the ellipse \(\displaystyle\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 40 square units
By symmetry, the rectangle has its corner in the first quadrant at a point \((5\cos\theta, 4\sin\theta)\) of the ellipse, \(0 \lt \theta \lt \frac{\pi}{2}\). Its sides are \(10\cos\theta\) and \(8\sin\theta\), so
\( A = 80\sin\theta\cos\theta = 40\sin 2\theta \)
Setting the derivative equal to zero:
\( \displaystyle \frac{dA}{d\theta} = 80\cos 2\theta = 0 \), so \( \displaystyle \theta = \frac{\pi}{4} \)
Since \(\frac{d^{2}A}{d\theta^{2}} = -160\sin 2\theta = -160 \lt 0\) there, this is a maximum:
\( A = 40\sin\frac{\pi}{2} = 40 \)
In general the largest such rectangle in \(\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1\) has area 2ab.
Why the other options are incorrect
Option (1) 20 square units — Incorrect. This is ab = 5 × 4, which is only half of the largest area 2ab.
Option (2) 80 square units — Incorrect. This is the area of the rectangle with sides 10 and 8 that encloses the ellipse; its corners (±5, ±4) are not on the ellipse.
Option (3) \(20\pi\) square units — Incorrect. This is the area of the ellipse itself, πab.