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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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Find the equation of an rectangular hyperbola whose centre is origin and its asymptotes coincide with the co-ordinate axes?

$\begin{array}{1 1}(a)\;xy=c\\(b)\;xy=c^2\\(c)\;x^2+y^2=a^2\\(d)\;x^2+y^2=1\end{array}$

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The equation of rectangular hyperbola is $x^2-y^2=a^2$ and its asymptotes are $x-y=0$ and $x+y=0$ .Since asymptotes are inclined at $45^{\large\circ}$ and 135$^{\large\circ}$ to the x-axis respectively.
If we rotate the axes through $\theta=-45^{\large\circ}$ without changing the origin.Thus when we replace (x,y) by
$[x\cos(-45^{\large\circ})-y\sin(-45^{\large\circ}),k\sin(-45^{\large\circ})+y\cos(-45^{\large\circ})]$
i.e $\big(\large\frac{x+y}{\sqrt y},\frac{-x+y}{\sqrt 2}\big)$
Then equation $x^2-y^2=a^2$ reduces to
$\big(\large\frac{x+y}{\sqrt{2}}\big)^2-\big(\large\frac{y-x}{\sqrt{2}}\big)^2$$=a^2$
$\large\frac{1}{2}$$(2y)(2x)=a^2$
$xy=\large\frac{a^2}{2}$$=c^2$
$xy=c^2$
Hence (b) is the correct answer.
answered Feb 10, 2014 by sreemathi.v
 

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