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

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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.

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