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(C) The equation of the parabola is $y^2 = 12x$,so $a = 3$. The tangent is $y = mx + \frac{3}{m}$.The equation of the hyperbola is $8x^2 - y^2 = 8$,wh

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$5 : 4$

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(C) The equation of the parabola is $y^2 = 12x$,so $a = 3$. The tangent is $y = mx + \frac{3}{m}$.The equation of the hyperbola is $8x^2 - y^2 = 8$,which simplifies to $x^2 - \frac{y^2}{8} = 1$. Here $a^2 = 1$ and $b^2 = 8$.The tangent is $y = mx \pm \sqrt{a^2m^2 - b^2} = mx \pm \sqrt{m^2 - 8}$.For common tangents,$\frac{3}{m} = \pm \sqrt{m^2 - 8}$.Squaring both sides,$\frac{9}{m^2} = m^2 - 8 \Rightarrow m^4 - 8m^2 - 9 = 0$.$(m^2 - 9)(m^2 + 1) = 0$. Since $m$ is real,$m^2 = 9$,so $m = \pm 3$.The common tangents are $y = 3x + 1$ and $y = -3x - 1$.Solving these,the intersection point $P$ is $(-1/3, 0)$.For the hyperbola $x^2 - \frac{y^2}{8} = 1$,$e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + 8} = 3$.The foci are $S(ae, 0) = (3, 0)$ and $S'(-ae, 0) = (-3, 0)$.Let $P$ divide $SS'$ in the ratio $k : 1$. Then $P = \left( \frac{k(-3) + 1(3)}{k+1}, 0 \right) = \left( \frac{3-3k}{k+1}, 0 \right)$.Equating to $P(-1/3, 0)$,we get $\frac{3-3k}{k+1} = -\frac{1}{3}$.$9 - 9k = -k - 1$ $\Rightarrow 8k = 10$ $\Rightarrow k = \frac{10}{8} = \frac{5}{4}$.Thus,the ratio is $5 : 4$.

Let $P$ be the point of intersection of the common tangents to the parabola $y^2 = 12x$ and the hyperbola $8x^2 - y^2 = 8$. If $S$ and $S'$ denote the foci of the hyperbola where $S$ lies on the positive $x$-axis,then $P$ divides $SS'$ in the ratio:

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