For what value of m is the line y = mx + 6 a | vedclass

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(B) The condition for the line $y = mx + c$ to be a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $c^2 = a^2m^2 - b^2$.Given the

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$\sqrt{\frac{17}{20}}$

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(B) The condition for the line $y = mx + c$ to be a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $c^2 = a^2m^2 - b^2$.Given the hyperbola $\frac{x^2}{100} - \frac{y^2}{49} = 1$,we have $a^2 = 100$ and $b^2 = 49$.The line is $y = mx + 6$,so $c = 6$.Substituting these values into the condition $c^2 = a^2m^2 - b^2$:$6^2 = 100m^2 - 49$$36 = 100m^2 - 49$$100m^2 = 36 + 49$$100m^2 = 85$$m^2 = \frac{85}{100} = \frac{17}{20}$$m = \sqrt{\frac{17}{20}}$.

For what value of $m$ is the line $y = mx + 6$ a tangent to the hyperbola $\frac{x^2}{100} - \frac{y^2}{49} = 1$?

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