The equation of common tangent of hyperbola $9 x^{2} - 9 y^{2} = 8$ and the parabola $y^{2} = 32 x$ is/are

9 x + 3 y - 8 = 0

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9 x - 3 y + 8 = 0

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9 x + 3 y + 8 = 0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

9 x - 3 y - 8 = 0

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Solution

The correct options are
B $9 x + 3 y + 8 = 0$
C $9 x - 3 y + 8 = 0$
Equation of tangent to the parabola $y^{2} = 32 x$ with slope $^{'} m^{'}$ is given by,
$y = m x + \frac{8}{m}$
Given it is also tangent to the hyperbola $x^{2} - y^{2} = \frac{8}{9}$
thus using condition of tangency, $c^{2} = a^{2} (m^{2} - 1)$
$\Rightarrow \frac{64}{m^{2}} = \frac{8}{9} (m^{2} - 1) \Rightarrow (m^{2} - 1) m^{2} = 72$
$\Rightarrow m^{2} = 9 \Rightarrow m = \pm 3$
Thus Equation of tangent are $y = 3 x + 8 / 3$ and $y = - 3 x - 8 / 3$
$\Rightarrow 9 x - 3 y + 8 = 0, 9 x + 3 y + 8 = 0$

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