logo

Ask Questions, Get Answers

X
 
Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class11  >>  Coordinate Geometry

Equation of the chord of the hyperbola $25x^2-16y^2=400$ which is bisected at the point $(6,2)$ is

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

1 Answer

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

Related questions

...