Equation of the two tangents drawn from $(2, - 1)$ to $x^{2} + 3 y^{2} = 3$ are:

x = 2, 4 x + 3 y - 7 = 0

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y = - 1, 4 x + y - 7 = 0

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x + y - 1 = 0, 4 x + y - 7 = 0

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x + y + 1 = 0, 4 x + y - 7 = 0

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Solution

The correct option is B $y = - 1, 4 x + y - 7 = 0$
Line passing through $(2, - 1)$ is

$y + 1 = m (x - 2)$

$y = m x + (- 1 - 2 m)$

Condition of tangency is $c^{2} = a^{2} m^{2} + b^{2}$

$(- 1 - 2 m)^{2} = 3 m^{2} + 1$

$1 + 4 m^{2} + 4 m = 3 m^{2} + 1$

$m^{2} + 4 m = 0 \Rightarrow m = 0, - 4$

$∴$ equation of tangents are $y = - 1$ and $4 x + y = 7$

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