logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
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
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...