Q97 / 160 · AP EAMCET 2019 (22 Apr Shift 1) · Mathematics
If \(l\) and \(b\) are respectively the length and breadth of the rectangle of greatest area that can be inscribed in the ellipse \(x^2+4 y^2=64\), then \((l, b)=\)
- A\((16 \sqrt{2}, 4 \sqrt{2})\)
- B\((8 \sqrt{2}, 6 \sqrt{2})\)
- Correct answerC\((8 \sqrt{2}, 4 \sqrt{2})\)
- D\((6 \sqrt{2}, 4 \sqrt{2})\)
Solution
Let \(P Q R S\) be a rectangle inscribed in the ellipse \(\frac{x^2}{64}+\frac{y^2}{16}=1\). Let the coordinate of \(P\) be \((8 \cos \theta, 4 \sin \theta)\). Then, the coordinates of \(Q, R\) and \(S\) are \((-8 \cos \theta, 4 \sin \theta),(-8 \cos \theta, 4 \sin \theta)\) and \((8 \cos \theta,-4 \sin \theta)\) respectively.
Let \(A\) be the area of rectangle \(P Q R S\). Then
\(\begin{gathered}
A=P Q \times R S \Rightarrow A=16 \cos \theta \times 8 \sin \theta \\
=64 \sin 20 \\
\Rightarrow \frac{d A}{d \theta}=128 \cos 2 \theta \text { and } \frac{d^2 A}{d \theta^2}=-256 \sin 2 \theta
\end{gathered}\)
The critical numbers of \(A\) are given by \(\frac{d A}{d \theta}=0\)
\(\therefore \quad \frac{d A}{d \theta}=0 \Rightarrow 4 a b \cos 2 \theta=0\)
\(\Rightarrow \quad \cos 2 \theta=0 \Rightarrow 2 \theta=\frac{\pi}{2}\) or \(\frac{3 \pi}{2}\)
\(\Rightarrow \quad \theta=\frac{\pi}{4}\) or \(\frac{3 \pi}{4}\)
Clearly, \(\left(\frac{d^2 A}{d \theta^2}\right)_{\theta=\frac{\pi}{4}}=-256 \sin \frac{\pi}{2}=-256 < 0\)
So, \(A\) is maximum when \(\theta=\frac{\pi}{4}\)
\(\therefore \quad P Q=l=16 \cos \theta=16 \cos \frac{\pi}{4}=16 \times \frac{1}{\sqrt{2}}=8 \sqrt{2}\)
and \(P S=b=8 \sin \theta=8 \sin \frac{\pi}{4}=8 \times \frac{1}{\sqrt{2}}=4 \sqrt{2}\)
Hence, \((l, b)=(8 \sqrt{2}, 4 \sqrt{2})\)
Hence, the correct answer is option: 'C'.