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Let the transverse axis of a hyperbola $\mathrm{H}$ be parallel to the $X-$ axis and $x^2+y^2-2 x-4 y+3=0$ be the equation of the auxiliary circle of $\mathrm{H}$. If the asymptotes of $\mathrm{H}$ are at right angles, then the equation of the hyperbola is
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The correct answer is:
$x^2-y^2-2 x+4 y-5=0$
Centre of auxiliary circle $=$ centre of hyperbola
$=(1,2)$ length of transverse axis = Diameter of auxiliary circle $=2 \mathrm{a}=2 \sqrt{1+4-3} \Rightarrow \mathrm{a}=\sqrt{2}$ A hyperbola whose asymptotes are at right angles to each other is rectangular hyperbola i.e. $\mathrm{e}=\sqrt{2}$
Since, $b^2=a^2\left(e^2-1\right)=2(2-1)=2 \Rightarrow b=\sqrt{2}$
Now, equation of hyperbola is
$\frac{(x-1)^2}{2}-\frac{(y-1)^2}{2}=1 \Rightarrow x^2-y^2-2 x+4 y-5=0$
$=(1,2)$ length of transverse axis = Diameter of auxiliary circle $=2 \mathrm{a}=2 \sqrt{1+4-3} \Rightarrow \mathrm{a}=\sqrt{2}$ A hyperbola whose asymptotes are at right angles to each other is rectangular hyperbola i.e. $\mathrm{e}=\sqrt{2}$
Since, $b^2=a^2\left(e^2-1\right)=2(2-1)=2 \Rightarrow b=\sqrt{2}$
Now, equation of hyperbola is
$\frac{(x-1)^2}{2}-\frac{(y-1)^2}{2}=1 \Rightarrow x^2-y^2-2 x+4 y-5=0$
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