Let m_1 and m_2 be the slopes of the tangents | vedclass

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

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$8$

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(C) The equation of a tangent to the hyperbola $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$ is $y = mx \pm \sqrt{a^2 - b^2m^2}$.Given the hyperbola $\frac{y^2}{25} - \frac{x^2}{16} = 1$,we have $a^2 = 25$ and $b^2 = 16$.For a tangent passing through $P(4, 1)$,we have $1 = 4m \pm \sqrt{25 - 16m^2}$.Rearranging gives $(1 - 4m)^2 = 25 - 16m^2$,which simplifies to $1 - 8m + 16m^2 = 25 - 16m^2$,or $32m^2 - 8m - 24 = 0$,which is $4m^2 - m - 3 = 0$.Solving for $m$,we get $(4m + 3)(m - 1) = 0$,so $m_1 = 1$ and $m_2 = -3/4$.The slopes for point $Q$ are $|m_1| = 1$ and $|m_2| = 3/4$.The tangent equations are $y = mx \pm \sqrt{25 - 16m^2}$. For positive $x$-intercepts,we choose the appropriate signs.For $m = 1$,$y = x \pm \sqrt{25 - 16} = x \pm 3$. The intercept is $x = -3$ or $x = 3$. To get a positive intercept,$y = x - 3$,so $\beta = 3$.For $m = 3/4$,$y = \frac{3}{4}x \pm \sqrt{25 - 16(9/16)} = \frac{3}{4}x \pm 4$. To get a positive intercept,$y = \frac{3}{4}x - 4$,so $\alpha = 16/3$.The intersection of $y = x - 3$ and $y = \frac{3}{4}x - 4$ gives $x - 3 = \frac{3}{4}x - 4$,so $\frac{1}{4}x = -1$,$x = -4$,and $y = -7$. Thus $Q = (-4, -7)$.$PQ^2 = (4 - (-4))^2 + (1 - (-7))^2 = 8^2 + 8^2 = 64 + 64 = 128$.Finally,$\frac{PQ^2}{\alpha \beta} = \frac{128}{(16/3) 	imes 3} = \frac{128}{16} = 8$.

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