If 3x + 2sqrt{2}y + k = 0 is a normal to the | vedclass

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(D) The equation of the hyperbola is $4x^2 - 9y^2 = 36$,which can be written as $\frac{x^2}{9} - \frac{y^2}{4} = 1$. Here,$a^2 = 9$ and $b^2 = 4$. The

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(D) The equation of the hyperbola is $4x^2 - 9y^2 = 36$,which can be written as $\frac{x^2}{9} - \frac{y^2}{4} = 1$. Here,$a^2 = 9$ and $b^2 = 4$. The normal to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at point $(x_0, y_0)$ is given by $\frac{a^2x}{x_0} + \frac{b^2y}{y_0} = a^2 + b^2$. Comparing this with the given normal $3x + 2\sqrt{2}y = -k$,we have $\frac{9}{x_0} = \frac{3}{\lambda}$ and $\frac{4}{y_0} = \frac{2\sqrt{2}}{\lambda}$,where $\lambda$ is a constant. Thus,$x_0 = 3\lambda$ and $y_0 = \frac{2}{\sqrt{2}}\lambda = \sqrt{2}\lambda$. Since $(x_0, y_0)$ lies on the hyperbola,$\frac{(3\lambda)^2}{9} - \frac{(\sqrt{2}\lambda)^2}{4} = 1$,which simplifies to $\lambda^2 - \frac{\lambda^2}{2} = 1$,so $\frac{\lambda^2}{2} = 1$,giving $\lambda^2 = 2$ or $\lambda = \pm\sqrt{2}$. The normal equation is $\frac{9x}{3\lambda} + \frac{4y}{\sqrt{2}\lambda} = 9 + 4 = 13$,so $\frac{3x}{\lambda} + \frac{2\sqrt{2}y}{\lambda} = 13$. Multiplying by $\lambda$,we get $3x + 2\sqrt{2}y = 13\lambda$. Comparing with $3x + 2\sqrt{2}y + k = 0$,we have $k = -13\lambda$. For positive intercepts on both axes,the line $3x + 2\sqrt{2}y = 13\lambda$ must have $13\lambda > 0$,so $\lambda = \sqrt{2}$. Thus,$k = -13\sqrt{2}$.

If $3x + 2\sqrt{2}y + k = 0$ is a normal to the hyperbola $4x^2 - 9y^2 - 36 = 0$ making positive intercepts on both the axes,then $k=$ (in $\sqrt{2}$)

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