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Q. The mid point of the chord $4x - 3y = 5 $ of the hyperbola $2x^2- 3y^2 - 12$ is

BITSATBITSAT 2009

Solution:

$4 x-3 y=5$
$2 x^{2}-3 y^{2}=12$
eliminate $y$ from line and put into the equation of hyperbola
$y=\left(\frac{4 x-5}{3}\right)$
$2 x^{2}-3\left(\frac{4 x-5}{3}\right)^{2}=12$
$x_{1}, x_{2}$ are the roots of the above quadratic
Now,
we have to find mid point of chord, if co-ordinates of chord of contact are $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$
so co-ordinates of mid point is $(x_1+x_2) / 2,(y_1+y_2) / 2$
$x_{1}+x_{2}=$ sum of roots $=4$
so $x-$ coordinate of mid point is $\frac{4}{2}=2$
put this $x-$ coordinates in chord we get $y=1$
so mid point is $( 2,1 )$