Let A(sec theta, 2 tan theta) and B(sec phi, | vedclass

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(D) The equation of the hyperbola is $2x^2 - y^2 = 2$,which can be written as $\frac{x^2}{1} - \frac{y^2}{2} = 1$. Here $a^2 = 1$ and $b^2 = 2$.The no

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(D) The equation of the hyperbola is $2x^2 - y^2 = 2$,which can be written as $\frac{x^2}{1} - \frac{y^2}{2} = 1$. Here $a^2 = 1$ and $b^2 = 2$.The normal at point $(x_1, y_1)$ on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $\frac{a^2x}{x_1} + \frac{b^2y}{y_1} = a^2 + b^2$.For point $A(\sec 	heta, 2 	an 	heta)$,the normal is $\frac{1 \cdot x}{\sec 	heta} + \frac{2 \cdot y}{2 	an 	heta} = 1 + 2 = 3$,which simplifies to $x \cos 	heta + y \cot 	heta = 3$.For point $B(\sec \phi, 2 	an \phi)$,the normal is $x \cos \phi + y \cot \phi = 3$.Given $\phi = \frac{\pi}{2} - 	heta$,we have $\cos \phi = \sin 	heta$ and $\cot \phi = 	an 	heta$.So the equations are:$1) x \cos 	heta + y \cot 	heta = 3$$2) x \sin 	heta + y 	an 	heta = 3$Solving for $y = \beta$ by eliminating $x$: Multiply $(1)$ by $\sin 	heta$ and $(2)$ by $\cos 	heta$:$x \cos 	heta \sin 	heta + y \cot 	heta \sin 	heta = 3 \sin 	heta$$x \sin 	heta \cos 	heta + y 	an 	heta \cos 	heta = 3 \cos 	heta$Subtracting the two equations:$y(\cos 	heta - \sin 	heta) = 3(\sin 	heta - \cos 	heta)$$y = -3$Thus,$\beta = -3$. Then $(2\beta)^2 = (2 	imes -3)^2 = (-6)^2 = 36$.

Let $A(\sec \theta, 2 \tan \theta)$ and $B(\sec \phi, 2 \tan \phi)$,where $\theta+\phi=\pi/2$,be two points on the hyperbola $2x^2-y^2=2$. If $(\alpha, \beta)$ is the point of intersection of the normals to the hyperbola at $A$ and $B$,then $(2\beta)^2$ is equal to ..... .

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