For the hyperbola frac{{{x^2}}}{9} - fra | vedclass

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Question

$p_1 p_2 = $ $\frac{{{a^2}\,{b^2}}}{{{a^2}\,\, + \,\,{b^2}}}$ ; $e = \sec 	heta$ 
$e^2 =1 +\frac{9}{3}\,$ $= 4$  $ \Rightarrow $ $e = 2 =

Vedclass · Accepted Answer

both $(A)$ and $(B)$

Vedclass · Answer

$p_1 p_2 = $ $\frac{{{a^2}\,{b^2}}}{{{a^2}\,\, + \,\,{b^2}}}$ ; $e = \sec 	heta$ 
$e^2 =1 +\frac{9}{3}\,$ $= 4$  $ \Rightarrow $ $e = 2 = \sec	heta $ ($B$  is correct)
$\Rightarrow$ $	heta = 60&deg;$
angle between the two asymptotes is $120&deg;$ 
$\Rightarrow $ acute angle is $60&deg;$ $ \Rightarrow $ $(A)$  is correct  
$C :$ $LLR =$ $\frac{{2{b^2}}}{a}\,$  = $2.\frac{3}{3}\,\,$ = $2 $ $\Rightarrow$  $(C)$ is correct  
$p_1p_2 $ = $\frac{{ab(\sec 	heta  + 	an 	heta )}}{{\sqrt {{a^2} + {b^2}} }}\,\,\,\frac{{ab(\sec 	heta  - 	an 	heta )}}{{\sqrt {{a^2} + {b^2}} }}$
$=$ $\frac{{{a^2}\,{b^2}}}{{{a^2} + {b^2}}}\,({\sec ^2}	heta  - {	an ^2}	heta )\,\, = \,\,\frac{{9\,.\,3}}{{12}}\,\, = \,\,\,\frac{9}{4}\,$

For the hyperbola $\frac{{{x^2}}}{9} - \frac{{{y^2}}}{3} = 1$ the incorrect statement is :

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