If the line 3x - my + 5 = 0 is a tangent to t | vedclass

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(D) Given the hyperbola equation: $3x^2 - 4y^2 = 300$. Dividing by $300$,we get: $\frac{x^2}{100} - \frac{y^2}{75} = 1$. Here,$a^2 = 100$ and $b^2 = 7

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$\frac{15}{7}$

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(D) Given the hyperbola equation: $3x^2 - 4y^2 = 300$. Dividing by $300$,we get: $\frac{x^2}{100} - \frac{y^2}{75} = 1$. Here,$a^2 = 100$ and $b^2 = 75$. The line is $3x - my + 5 = 0$,which can be written as $my = 3x + 5$,or $y = \frac{3}{m}x + \frac{5}{m}$. Comparing with $y = Mx + c$,we have $M = \frac{3}{m}$ and $c = \frac{5}{m}$. 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 $c^2 = a^2M^2 - b^2$. Substituting the values: $(\frac{5}{m})^2 = 100(\frac{3}{m})^2 - 75$. $\frac{25}{m^2} = \frac{900}{m^2} - 75$. $75 = \frac{900 - 25}{m^2} = \frac{875}{m^2}$. $m^2 = \frac{875}{75} = \frac{35}{3}$. The $Y$-intercept is $c = \frac{5}{m}$,so the square of the $Y$-intercept is $c^2 = \frac{25}{m^2}$. Substituting $m^2 = \frac{35}{3}$: $c^2 = 25 	imes \frac{3}{35} = \frac{5 	imes 3}{7} = \frac{15}{7}$.

If the line $3x - my + 5 = 0$ is a tangent to the hyperbola $3x^2 - 4y^2 = 300$,then the square of the $Y$-intercept made by this tangent line is:

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