Let P be the point of intersection of the com | vedclass

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Question

Tangents ${y^2} = 12x \Rightarrow y = 2x + \frac{3}{m}$

$\frac{{{x^2}}}{1} - \frac{{{y^2}}}{8} = 1 \Rightarrow y = mx \pm \sqrt {{m^2} - 8} $

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$5 : 4$

Vedclass · Answer

Tangents ${y^2} = 12x \Rightarrow y = 2x + \frac{3}{m}$

$\frac{{{x^2}}}{1} - \frac{{{y^2}}}{8} = 1 \Rightarrow y = mx \pm \sqrt {{m^2} - 8} $

Common tangent given

$	herefore \frac{3}{{m =  \pm \sqrt {{m^2} - 8} }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{{x^2}}}{1} - \frac{{{y^2}}}{8} = 1$

${m^4} - 8{m^2} - 9 = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,e = 3$

$m =  \pm 3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,ae = 3$

$	herefore y = 3x + 1\,\,\,\,\,P\left( { - \frac{1}{3},0} ight)\,\,\,\,\,S = \left( {3,0} ight)$

$y =  - 3x - 1$ $P$ divies $SS'$ in $5:4$.

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 a ratio

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