If 3 sqrt{2} x - 4 y = 12 is a tangent to the | vedclass

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(B) The equation of the tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ with slope $m$ is $y = mx \pm \sqrt{a^2 m^2 - b^2}$.Given the

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(B) The equation of the tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ with slope $m$ is $y = mx \pm \sqrt{a^2 m^2 - b^2}$.Given the tangent line $3 \sqrt{2} x - 4 y = 12$,we rewrite it as $y = \frac{3 \sqrt{2}}{4} x - 3$. Thus,$m = \frac{3 \sqrt{2}}{4}$ and the constant term is $-3$,so $\sqrt{a^2 m^2 - b^2} = 3$,which implies $a^2 m^2 - b^2 = 9$.Substituting $m^2 = \frac{18}{16} = \frac{9}{8}$,we get $\frac{9}{8} a^2 - b^2 = 9$.Given eccentricity $e = \frac{5}{4}$,we know $e^2 = 1 + \frac{b^2}{a^2}$,so $\frac{25}{16} = 1 + \frac{b^2}{a^2}$,which gives $\frac{b^2}{a^2} = \frac{9}{16}$,or $b^2 = \frac{9}{16} a^2$.Substituting $b^2$ into the tangent condition: $\frac{9}{8} a^2 - \frac{9}{16} a^2 = 9$.Multiplying by $16$: $18 a^2 - 9 a^2 = 144$,so $9 a^2 = 144$,which means $a^2 = 16$.Then $b^2 = \frac{9}{16} 	imes 16 = 9$.Finally,$a^2 - b^2 = 16 - 9 = 7$.

If $3 \sqrt{2} x - 4 y = 12$ is a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ and $\frac{5}{4}$ is its eccentricity,then $a^2 - b^2 =$

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