If the pole of the line 3x - 16y + 48 = 0 wit | vedclass

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(C) The equation of the polar of a point $(\alpha, \beta)$ with respect to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is given by $\frac{\a

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$2$

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(C) The equation of the polar of a point $(\alpha, \beta)$ with respect to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is given by $\frac{\alpha x}{a^2} - \frac{\beta y}{b^2} = 1$. Given hyperbola is $9x^2 - 16y^2 = 144$,which can be written as $\frac{x^2}{16} - \frac{y^2}{9} = 1$. So,the equation of the polar of $(\alpha, \beta)$ is $\frac{\alpha x}{16} - \frac{\beta y}{9} = 1$,or $9\alpha x - 16\beta y - 144 = 0$. Comparing this with the given line $3x - 16y + 48 = 0$,we have: $\frac{9\alpha}{3} = \frac{-16\beta}{-16} = \frac{-144}{48}$. $3\alpha = \beta = -3$. Thus,$\alpha = -1$ and $\beta = -3$. Therefore,$\alpha - \beta = -1 - (-3) = -1 + 3 = 2$.

If the pole of the line $3x - 16y + 48 = 0$ with respect to the hyperbola $9x^2 - 16y^2 = 144$ is $(\alpha, \beta)$,then $\alpha - \beta = $

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