Let S be the focus of the hyperbola x^2 - 2y^ | vedclass

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(C) The equation of the hyperbola is $x^2 - 2y^2 = 1$. Comparing with $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we get $a^2 = 1$ and $b^2 = \frac{1}{2}$

Vedclass · Accepted Answer

$\frac{3}{2(2+\sqrt{6})}$

Vedclass · Answer

(C) The equation of the hyperbola is $x^2 - 2y^2 = 1$. Comparing with $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we get $a^2 = 1$ and $b^2 = \frac{1}{2}$.The eccentricity $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{1/2}{1}} = \sqrt{\frac{3}{2}}$.The focus $S$ is $(ae, 0) = (\sqrt{\frac{3}{2}}, 0)$.The line passing through $P(-1, 1)$ and $S(\sqrt{\frac{3}{2}}, 0)$ has slope $m = \frac{0 - 1}{\sqrt{\frac{3}{2}} - (-1)} = \frac{-1}{\sqrt{\frac{3}{2}} + 1} = \frac{-\sqrt{2}}{\sqrt{3} + \sqrt{2}}$.The equation of line $PS$ is $y - 0 = m(x - \sqrt{\frac{3}{2}})$,which is $y = \frac{-\sqrt{2}}{\sqrt{3} + \sqrt{2}}(x - \sqrt{\frac{3}{2}})$.To find the $Y$-intercept $(B)$,set $x = 0$: $y = \frac{-\sqrt{2}}{\sqrt{3} + \sqrt{2}}(-\sqrt{\frac{3}{2}}) = \frac{\sqrt{3}}{\sqrt{3} + \sqrt{2}}$.The area of the triangle formed by the line with coordinate axes is $\frac{1}{2} 	imes |OS| 	imes |OB| = \frac{1}{2} 	imes \sqrt{\frac{3}{2}} 	imes \frac{\sqrt{3}}{\sqrt{3} + \sqrt{2}} = \frac{1}{2} 	imes \frac{3}{\sqrt{2}(\sqrt{3} + \sqrt{2})} = \frac{3}{2(\sqrt{6} + 2)}$.Thus,the correct option is $C$.

Let $S$ be the focus of the hyperbola $x^2 - 2y^2 = 1$ lying on the positive $X$-axis. Let $P(-1, 1)$ be a given point. Then the area of the triangle formed by the line $PS$ with the coordinate axes is (in sq. units)

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