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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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Equation of chord of the hyperbola $25x^2-16y^2=400$ which is bisected at the point $(6,2)$ is

$\begin{array}{1 1}(A)\;a\;16x-75y=418 \\(B)\;75x-16y=418 \\(C)\;25x-4y=400 \\(D)\;none\;of\;these \end{array}$

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1 Answer

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$25x^2-16y^2=400$
$\large\frac{x^2}{16}-\frac{y^2}{25}$$=1$
Chord bisected at $(6,2)$
$\large\frac{6x}{16}-\frac{2y}{25} =\frac{6^2}{16}-\frac{2^2}{25}$
=> $25 \times 6x -16 \times 2y =25 \times 36 -16 \times 4$
=>$ 150 x -32 y =900 -64$
i.e $75x - 16 y=418$
Hence B is the correct answer.
answered Apr 21, 2014 by meena.p
 

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