Find the equation of the tangents drawn from the point $(- 2, - 1)$ to the hyperbola $2 x^{2} - 3 y^{2} = 6.$

x - y + 1 = 0

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3 x - y - 7 = 0

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3 x - y + 7 = 0

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x + y + 1 = 0

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Solution

The correct options are
A $x - y + 1 = 0$
B $3 x - y + 7 = 0$
Any tangent is $y = m x \pm \sqrt{(a^{2} m^{2} - b^{2})}$
It passes through $(- 2, - 1)$
$∴ {(2 m - 1)}^{2} = 3 m^{2} - 2$ or $m^{2} - 4 m + 3 = 0$
$∴ m = 3$ or $1$ .
Hence, the tangents on substituting $m$ are $3 x - y + 7 = 0, x - y + 1 = 0$

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