If the tangent at the point (2 sec phi, 3 tan | vedclass

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(C) The equation of the hyperbola is $\frac{x^2}{4} - \frac{y^2}{9} = 1$. Given the point $(x, y) = (2 \sec \phi, 3 	an \phi)$. Differentiating $x =

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$30$

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(C) The equation of the hyperbola is $\frac{x^2}{4} - \frac{y^2}{9} = 1$. Given the point $(x, y) = (2 \sec \phi, 3 	an \phi)$. Differentiating $x = 2 \sec \phi$ with respect to $\phi$,we get $\frac{dx}{d\phi} = 2 \sec \phi 	an \phi$. Differentiating $y = 3 	an \phi$ with respect to $\phi$,we get $\frac{dy}{d\phi} = 3 \sec^2 \phi$. The slope of the tangent is $\frac{dy}{dx} = \frac{dy/d\phi}{dx/d\phi} = \frac{3 \sec^2 \phi}{2 \sec \phi 	an \phi} = \frac{3 \sec \phi}{2 	an \phi} = \frac{3}{2 \cos \phi} \cdot \frac{\cos \phi}{\sin \phi} = \frac{3}{2 \sin \phi} = \frac{3}{2} \csc \phi$. The given line is $3x - y + 4 = 0$,which can be written as $y = 3x + 4$. Its slope is $3$. Since the tangent is parallel to the line,their slopes are equal: $\frac{3}{2} \csc \phi = 3$. This implies $\csc \phi = 2$,so $\sin \phi = \frac{1}{2}$. Thus,$\phi = 30^\circ$.

If the tangent at the point $(2 \sec \phi, 3 \tan \phi)$ of the hyperbola $\frac{x^2}{4} - \frac{y^2}{9} = 1$ is parallel to $3x - y + 4 = 0$,then the value of $\phi$ is ............ $^o$.

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