If the line 5x - 2y - 6 = 0 is a tangent to t | vedclass

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(C) The given equation of the hyperbola is $5x^2 - ky^2 = 12$,which can be written as $\frac{x^2}{12/5} - \frac{y^2}{12/k} = 1$.Comparing this with $\

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$\sqrt{6}x - 5y = 21$

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(C) The given equation of the hyperbola is $5x^2 - ky^2 = 12$,which can be written as $\frac{x^2}{12/5} - \frac{y^2}{12/k} = 1$.Comparing this with $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we have $a^2 = \frac{12}{5}$ and $b^2 = \frac{12}{k}$.The condition for the line $y = mx + c$ to be a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $c^2 = a^2m^2 - b^2$.Rewriting the line $5x - 2y - 6 = 0$ as $y = \frac{5}{2}x - 3$,we have $m = \frac{5}{2}$ and $c = -3$.Substituting these values: $(-3)^2 = \frac{12}{5} 	imes (\frac{5}{2})^2 - \frac{12}{k}$ $\Rightarrow 9 = \frac{12}{5} 	imes \frac{25}{4} - \frac{12}{k}$ $\Rightarrow 9 = 15 - \frac{12}{k}$ $\Rightarrow \frac{12}{k} = 6$ $\Rightarrow k = 2$.Now,the hyperbola is $5x^2 - 2y^2 = 12$. The point $(\sqrt{6}, p)$ lies on it,so $5(\sqrt{6})^2 - 2p^2 = 12$ $\Rightarrow 30 - 2p^2 = 12$ $\Rightarrow 2p^2 = 18$ $\Rightarrow p^2 = 9$. Since $p The equation of the normal to $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at $(x_1, y_1)$ is $\frac{a^2x}{x_1} + \frac{b^2y}{y_1} = a^2 + b^2$.Here $a^2 = \frac{12}{5}$,$b^2 = \frac{12}{2} = 6$,$x_1 = \sqrt{6}$,$y_1 = -3$.Substituting these: $\frac{(12/5)x}{\sqrt{6}} + \frac{6y}{-3} = \frac{12}{5} + 6$ $\Rightarrow \frac{2\sqrt{6}x}{5} - 2y = \frac{42}{5}$ $\Rightarrow 2\sqrt{6}x - 10y = 42$ $\Rightarrow \sqrt{6}x - 5y = 21$.

If the line $5x - 2y - 6 = 0$ is a tangent to the hyperbola $5x^2 - ky^2 = 12$,then the equation of the normal to this hyperbola at the point $(\sqrt{6}, p)$ where $p < 0$ is:

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