Equation of the normal to the hyperbola $3 x^{2} - y^{2} = 3 a t (2, - 3)$ is

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$3 x + 2 y - 14 = 0$

$x + 2 y + 4 = 0$

$x - 2 y - 8 = 0$

$3 x - 2 y - 12 = 0$

answer is A.

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Detailed Solution

$\frac{a^{2} x}{x_{1}} + \frac{b^{2} y}{y_{1}} = a^{2} + b^{2}$

$\frac{x^{2}}{1} - \frac{y^{2}}{3} = 1$

$\Rightarrow a^{2} = 1, b^{2} = 3$ , $(x_{1}, y_{1}) = (2, - 3)$

$\frac{a^{2} x}{x_{1}} + \frac{b^{2} y}{y_{1}} = a^{2} + b^{2}$

$\Rightarrow \frac{1. x}{2} + \frac{3. y}{- 3} = 1 + 3 \Rightarrow \frac{x}{2} - y = 4 \Rightarrow x - 2 y = 8$

$\Rightarrow x - 2 y - 8 = 0$

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