If the product of the perpendicular distances | vedclass

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Question

$p_1p_2 $ =$\frac{{{a^2}{b^2}}}{{{a^2} + {b^2}}}\,$ = $\frac{{{a^2}.\,{a^2}\,({e^2} - 1)}}{{{a^2}\,{e^2}}}\,\,$ $= 6$
$\frac{{2{a^2}}}{3}\,\, = \,\,6

Vedclass · Accepted Answer

$6$

Vedclass · Answer

$p_1p_2 $ =$\frac{{{a^2}{b^2}}}{{{a^2} + {b^2}}}\,$ = $\frac{{{a^2}.\,{a^2}\,({e^2} - 1)}}{{{a^2}\,{e^2}}}\,\,$ $= 6$
$\frac{{2{a^2}}}{3}\,\, = \,\,6\,$ $ \Rightarrow$ $ a^2 = 9 $ $\Rightarrow $ $a = 3$
hence $ 2a = 6$

If the product of the perpendicular distances from any point on the hyperbola $\frac{{{x^2}}}{{{a^2}}}\,\, - \,\,\frac{{{y^2}}}{{{b^2}}}\,\,\, = \,1$ of eccentricity $e =\sqrt 3 \,$ from its asymptotes is equal to $6$, then the length of the transverse axis of the hyperbola is

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