A hyperbola having the transverse axis of len | vedclass

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(B) The given ellipse is $3x^{2} + 4y^{2} = 12$,which can be written as $\frac{x^{2}}{4} + \frac{y^{2}}{3} = 1$.Here,$a^{2} = 4$ and $b^{2} = 3$. The

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$\left(\sqrt{\frac{3}{2}}, \frac{1}{\sqrt{2}}\right)$

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(B) The given ellipse is $3x^{2} + 4y^{2} = 12$,which can be written as $\frac{x^{2}}{4} + \frac{y^{2}}{3} = 1$.Here,$a^{2} = 4$ and $b^{2} = 3$. The eccentricity $e = \sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{1 - \frac{3}{4}} = \frac{1}{2}$.The foci are $(\pm ae, 0) = (\pm 2 	imes \frac{1}{2}, 0) = (\pm 1, 0)$.For the hyperbola,the length of the transverse axis is $2a_{h} = \sqrt{2}$,so $a_{h} = \frac{1}{\sqrt{2}}$ and $a_{h}^{2} = \frac{1}{2}$.Let the hyperbola be $\frac{x^{2}}{a_{h}^{2}} - \frac{y^{2}}{b_{h}^{2}} = 1$,i.e.,$\frac{x^{2}}{1/2} - \frac{y^{2}}{b_{h}^{2}} = 1$.The foci of the hyperbola are $(\pm a_{h}e_{h}, 0)$,where $e_{h} = \sqrt{1 + \frac{b_{h}^{2}}{a_{h}^{2}}} = \sqrt{1 + 2b_{h}^{2}}$.Thus,the foci are $(\pm \sqrt{a_{h}^{2} + b_{h}^{2}}, 0) = (\pm \sqrt{\frac{1}{2} + b_{h}^{2}}, 0)$.Since the foci are the same,$\sqrt{\frac{1}{2} + b_{h}^{2}} = 1$,which implies $\frac{1}{2} + b_{h}^{2} = 1$,so $b_{h}^{2} = \frac{1}{2}$.The equation of the hyperbola is $\frac{x^{2}}{1/2} - \frac{y^{2}}{1/2} = 1$,or $x^{2} - y^{2} = \frac{1}{2}$.Checking the points:For option $B$: $(\sqrt{3/2})^{2} - (1/\sqrt{2})^{2} = \frac{3}{2} - \frac{1}{2} = 1 
eq \frac{1}{2}$.Thus,the point $\left(\sqrt{\frac{3}{2}}, \frac{1}{\sqrt{2}}ight)$ does not lie on the hyperbola.

$A$ hyperbola having the transverse axis of length $\sqrt{2}$ has the same foci as that of the ellipse $3x^{2} + 4y^{2} = 12$. Then this hyperbola does not pass through which of the following points?

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