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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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Find the equation of a diameter bisecting a system of parallel chords of slope m of the hyperbola $\large\frac{x^2}{a^2}-\frac{y^2}{b^2}$$=1$?

$\begin{array}{1 1}(a)\;y=\large\frac{b^2x}{a^2m}\\(b)\;y=\large\frac{a^2x}{b^2m}\\(c)\;y=\large\frac{bx}{am}\\(d)\;\text{None of these}\end{array}$

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Let $y=mx+c$ be a system of a parallel chords to $\large\frac{x^2}{a^2}-\frac{y^2}{b^2}$$=1$ for different chords
(c varies)
Let $(x_1,y_1)$ & $(x_2,y_2)$ be the extremities of chord and (h,k) be its middle point.
The $\large\frac{x^2}{a^2}-\frac{y^2}{b^2}$$=1$ and $y=mx+c$
$\large\frac{x^2}{a^2}-\frac{(mx+c)^2}{b^2}$$=1$
$(a^2m^2-b^2)x^2+2mca^2x+a^2(c^2+b^2)=0$
$x_1$ & $x_2$ be the roots
$x_1+x_2=-\large\frac{2mca^2}{a^2m^2b^2}$
$(h,k)$ is middle point hence
$x_1+x_2=2h$
$2h=-\large\frac{2mca^2}{a^2m^2b^2}$
$(h,k)$ lies on $y=mx+c$
Hence $c=k-mh$
$h=-\large\frac{-m(k-mh)a^2}{a^2m^2-b^2}$
$k=\large\frac{b^2h}{a^2m}$
Hence $y=\large\frac{b^2x}{a^2m}$
Hence (a) is the correct answer.
answered Feb 10, 2014 by sreemathi.v
 

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