Let the transverse axis of a hyperbola H be p | vedclass

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(D) The equation of the auxiliary circle is $x^2+y^2-2x-4y+3=0$. Rewriting it as $(x-1)^2+(y-2)^2 = 1+4-3 = 2$. The centre of the hyperbola is the sam

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$x^2-y^2-2x+4y-5=0$

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(D) The equation of the auxiliary circle is $x^2+y^2-2x-4y+3=0$. Rewriting it as $(x-1)^2+(y-2)^2 = 1+4-3 = 2$. The centre of the hyperbola is the same as the centre of the auxiliary circle,which is $(1, 2)$. The radius of the auxiliary circle is $a = \sqrt{2}$. Since the asymptotes are at right angles,the hyperbola is a rectangular hyperbola,so $e = \sqrt{2}$. For a rectangular hyperbola,$a = b$,so $b = \sqrt{2}$. The equation of the hyperbola with transverse axis parallel to the $X$-axis is $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$. Substituting the values: $\frac{(x-1)^2}{2} - \frac{(y-2)^2}{2} = 1$. $(x-1)^2 - (y-2)^2 = 2$. $x^2-2x+1 - (y^2-4y+4) = 2$. $x^2-y^2-2x+4y-3-2 = 0$. $x^2-y^2-2x+4y-5 = 0$.

Let the transverse axis of a hyperbola $H$ be parallel to the $X$-axis and $x^2+y^2-2x-4y+3=0$ be the equation of the auxiliary circle of $H$. If the asymptotes of $H$ are at right angles,then the equation of the hyperbola is

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