Let the focal chord of the parabola P: y^{2}= | vedclass

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(B) The parabola is $P: y^{2}=4x$,so its focus is $(1, 0)$. Since $L: y=mx+c$ is a focal chord,it passes through $(1, 0)$,so $0 = m(1) + c$,which give

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$2\sqrt{14}$

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(B) The parabola is $P: y^{2}=4x$,so its focus is $(1, 0)$. Since $L: y=mx+c$ is a focal chord,it passes through $(1, 0)$,so $0 = m(1) + c$,which gives $c = -m$.Thus,the line is $y = m(x-1)$.The hyperbola is $H: x^{2}-y^{2}=4$,or $\frac{x^{2}}{4} - \frac{y^{2}}{4} = 1$. Here $a^{2}=4, b^{2}=4$.The condition for $y=mx+c$ to be a tangent to the hyperbola is $c^{2} = a^{2}m^{2} - b^{2}$.Substituting $c = -m$,we get $(-m)^{2} = 4m^{2} - 4$,so $m^{2} = 4m^{2} - 4$,which implies $3m^{2} = 4$,or $m = \frac{2}{\sqrt{3}}$ (since $m>0$).Then $c = -\frac{2}{\sqrt{3}}$. The line $L$ is $y = \frac{2}{\sqrt{3}}(x-1)$.The intersection points $M(x_{1}, y_{1})$ and $N(x_{2}, y_{2})$ are found by substituting $y = \frac{2}{\sqrt{3}}(x-1)$ into $y^{2}=4x$:$\frac{4}{3}(x-1)^{2} = 4x \implies (x-1)^{2} = 3x \implies x^{2}-2x+1 = 3x \implies x^{2}-5x+1 = 0$.The roots are $x_{1}, x_{2}$,so $x_{1}+x_{2}=5$ and $x_{1}x_{2}=1$.The $y$-coordinates are $y_{i} = \frac{2}{\sqrt{3}}(x_{i}-1)$.The area of quadrilateral $OMFN$ is the sum of areas of $	riangle OMF$ and $	riangle ONF$.Area $= \frac{1}{2} |x_{F} y_{1} - x_{F} y_{2}| = \frac{1}{2} |x_{F}| |y_{1}-y_{2}|$.Here $x_{F} = 2\sqrt{2}$ (focus of $H$ is $(ae, 0) = (2\sqrt{2}, 0)$).$|y_{1}-y_{2}| = \frac{2}{\sqrt{3}} |x_{1}-x_{2}| = \frac{2}{\sqrt{3}} \sqrt{(x_{1}+x_{2})^{2}-4x_{1}x_{2}} = \frac{2}{\sqrt{3}} \sqrt{25-4} = \frac{2\sqrt{21}}{\sqrt{3}} = 2\sqrt{7}$.Area $= \frac{1}{2} (2\sqrt{2}) (2\sqrt{7}) = 2\sqrt{14}$.

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