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A rectangular hyperbola passing through $(3,2)$ has its asymptotes parallel to the coordinate axes. If $(1,1)$ is the point of intersection of the two perpendicular tangents of that hyperbola, then its equation is
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The correct answer is:
$x y=x+y+1$
It is given that a rectangular hyperbola whose asymptotes are parallel to coordinate axis and point of intersection of perpendicular tangents (means centre of hyperbola) is $(1,1)$, the equation of hyperbola we can take as
$(x-1)(y-1)=k \text {, }$ $\ldots(\mathrm{i})$
$\because$ Hyperbola (i) passes through point $(3,2)$, so $k=2$
Therefore equation of the required hyperbola is
$(x-1)(y-1)=2 \Rightarrow x y=x+y+1$
$(x-1)(y-1)=k \text {, }$ $\ldots(\mathrm{i})$
$\because$ Hyperbola (i) passes through point $(3,2)$, so $k=2$
Therefore equation of the required hyperbola is
$(x-1)(y-1)=2 \Rightarrow x y=x+y+1$
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