Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Model Papers
0 votes

Find the area of the greatest rectangle that can be inscribed in an ellipse $\Large \frac{x^2}{a^2}+\frac{y^2}{b^2}\normalsize = 1.$

(Note: This question has been split into 2 questions) This question appeared in 65-1,65-2 and 65-3 versions of the paper in 2013.
Can you answer this question?

1 Answer

0 votes
$\textbf{Step 1}$:
Let $(a\cos \theta.b \sin \theta)$ be one of the vertices of the rectangle.
Hence the length=$2a \cos \theta$ and the width=$2b \sin \theta$
$\Rightarrow$ Area of the rectangle $A\; = 2a\cos \theta \times 2 b \sin \theta$ $=2ab.2 \sin \theta \cos \theta$
$\Rightarrow A =2ab\sin 2 \theta$
$\Rightarrow A'=2ab.2\cos 2 \theta$
$\textbf{Step 2}$:
For maximun or minimum values of A, $\rightarrow A'=0$
$\Rightarrow 4ab\cos 2 \theta=0$
$\Rightarrow \cos 2 \theta=\large \frac{\pi}{2}$ $\rightarrow \theta =\frac{\pi}{4}$
$\Rightarrow A =2ab \sin2\frac{\pi}{4}$
$\Rightarrow A=2ab \sin (\large\frac{\pi}{2})$$=2ab$
Now $A''=-8ab \sin 2 \theta \rightarrow \;$Area is maximum when $A=2ab$
answered Mar 22, 2013 by balaji.thirumalai
edited Apr 2, 2013 by meena.p

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App