If the tangent at the point (2 sec theta, 3 t | vedclass

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(C) The equation of the hyperbola is $\frac{x^2}{4}-\frac{y^2}{9}=1$. Comparing this with $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$,we get $a=2$ and $b=3$.

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(C) The equation of the hyperbola is $\frac{x^2}{4}-\frac{y^2}{9}=1$. Comparing this with $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$,we get $a=2$ and $b=3$. The equation of the tangent at the point $(a \sec 	heta, b 	an 	heta)$ is $\frac{x \sec 	heta}{a} - \frac{y 	an 	heta}{b} = 1$. Substituting $a=2$ and $b=3$,the tangent equation is $\frac{x \sec 	heta}{2} - \frac{y 	an 	heta}{3} = 1$. This can be rewritten as $y = \left( \frac{3 \sec 	heta}{2 	an 	heta} ight) x - \frac{3}{	an 	heta}$. The slope of this tangent is $m_1 = \frac{3 \sec 	heta}{2 	an 	heta} = \frac{3}{2 \sin 	heta}$. The given line is $3x-y+4=0$,which can be written as $y=3x+4$. The slope of this line is $m_2 = 3$. Since the tangent is parallel to the line,$m_1 = m_2$. $\frac{3}{2 \sin 	heta} = 3 \implies \sin 	heta = \frac{1}{2}$. Thus,$	heta = 30^{\circ}$.

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

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