Point 'O' is the centre of the ellip | vedclass

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Question

$a^2 e^2 = 36 \Rightarrow a^2 - b^2 = 36..... (1)$ 
Using $r = (s - a) 	an \frac {A}{2}\, in\, \Delta OCF $ 
$1 = (s - a) tan 45&deg;$ wh

Vedclass · Accepted Answer

$65$

Vedclass · Answer

$a^2 e^2 = 36 \Rightarrow a^2 - b^2 = 36..... (1)$ 
Using $r = (s - a) 	an \frac {A}{2}\, in\, \Delta OCF $ 
$1 = (s - a) tan 45&deg;$ when $a = CF$ 
$2 = 2 (s - a)$ 
$= 2s - 2a = 2s - AB $ 
$= (OF + FC + CO) - AB $ 
$2 = 6 +\frac{{A\,B}}{2} + \frac{{C\,D}}{2} - AB$ 
$ \frac{AB-CD}{2} = 4 \Rightarrow 2 (a - b) = 8 \Rightarrow a - b = 4.... (2)$ 
From $(1)$ and $ (2)$ $ a + b = 9$

$ \Rightarrow $ $2a = 13$  $;$ $ 2b = 5$

$\Rightarrow $ $(AB) (CD) = 65$

Point $'O' $ is the centre of the ellipse with major axis $AB$ $ \&$ minor axis $CD$. Point $F$ is one focus of the ellipse. If $OF = 6 $ $ \&$ the diameter of the inscribed circle of triangle $OCF$ is $2, $ then the product $ (AB)\,(CD) $ is equal to

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