A rectangular hyperbola of latus rectum 2 uni | vedclass

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(B) For a rectangular hyperbola,the eccentricity $e = \sqrt{2}$.The length of the latus rectum is $\frac{2b^2}{a} = 2$,which implies $b^2 = a$.For a r

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$x^2 + y^2 = 9$

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(B) For a rectangular hyperbola,the eccentricity $e = \sqrt{2}$.The length of the latus rectum is $\frac{2b^2}{a} = 2$,which implies $b^2 = a$.For a rectangular hyperbola,$a^2 = b^2$,so $a^2 = a$,which gives $a = 1$ (since $a > 0$).Thus,$a = 1$ and $b = 1$.The distance between the foci is $2ae = 2(1)(\sqrt{2}) = 2\sqrt{2}$.Let the other focus be $(h, k)$. The given focus is $S_1 = (1, 0)$ and the point on the hyperbola is $P = (0, 0)$.By the definition of a hyperbola,$|PS_1 - PS_2| = 2a$.$|\sqrt{(0-1)^2 + (0-0)^2} - \sqrt{(0-h)^2 + (0-k)^2}| = 2(1)$.$|1 - \sqrt{h^2 + k^2}| = 2$.This implies $\sqrt{h^2 + k^2} = 3$ or $\sqrt{h^2 + k^2} = -1$ (impossible).Therefore,$h^2 + k^2 = 9$,which represents the curve $x^2 + y^2 = 9$.

$A$ rectangular hyperbola of latus rectum $2$ units passes through $(0, 0)$ and has $(1, 0)$ as its one focus. The other focus lies on the curve -

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