$T h e l e n g t h o f t h e s t r a i g h t l i n e x - 3 y = 1 i n t e r c e p t e d b y t h e h y p e r b o l a x^{2} - 4 y^{2} = 1 i s$

$\frac{6}{\sqrt{5}}$

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$3 \sqrt{\frac{2}{5}}$

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$6 \sqrt{\frac{2}{5}}$

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none of these

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Solution

The correct option is C
$6 \sqrt{\frac{2}{5}}$

$T h e p o i n t o f i n t e r s e c t i o n o f x - 3 y = 1$

$a n d t h e h y p e r b o l a x^{2} - 14 y^{2} = 1 i s$

$c a l c u l a t e d i n t h e f o l l o w i n g w a y :$

$(1 + 3 y)^{2} - 4 y^{2} = 1$

$\Rightarrow 1 + 6 y + 9 y^{2} - 4 y^{2} = 1$

$\Rightarrow 5 y^{2} + 6 y = 0$

$\Rightarrow y = 0 o r y = - \frac{6}{5}$

$I f Y = 0 t h e n x = 1$

$I f y = - \frac{6}{5} t h e n x = 1 + 3 \times (- \frac{6}{5}) = - \frac{13}{5}$

$S o, t h e p o i n t s a r e (1, 0) a n d (- \frac{13}{5}, - \frac{6}{5})$

$∴ L e n g t h = \sqrt{{(1 + \frac{13}{5})}^{2} + {(0 + \frac{6}{5})}^{2}} = 6 \sqrt{\frac{2}{5}}$

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Similar questions

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The length of the straight line x−3y=1 intercepted by the hyperbola x2−4y2=1 is

$T h e l e n g t h o f t h e s t r a i g h t l i n e x - 3 y = 1 i n t e r c e p t e d b y t h e h y p e r b o l a x^{2} - 4 y^{2} = 1 i s$