Let A(2 sec theta, 3 tan theta) and B(2 sec p | vedclass

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(A) The equation of the normal to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at point $(a \sec 	heta, b 	an 	heta)$ is $ax \cos 	heta +

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$-\frac{13}{3}$

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(A) The equation of the normal to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at point $(a \sec 	heta, b 	an 	heta)$ is $ax \cos 	heta + by \cot 	heta = a^2 + b^2$.For the given hyperbola $\frac{x^2}{4} - \frac{y^2}{9} = 1$,we have $a^2 = 4$ and $b^2 = 9$.The normal at $A(2 \sec 	heta, 3 	an 	heta)$ is $2x \cos 	heta + 3y \cot 	heta = 4 + 9 = 13$ $(i)$.The normal at $B(2 \sec \phi, 3 	an \phi)$ is $2x \cos \phi + 3y \cot \phi = 4 + 9 = 13$ (ii).Since $	heta + \phi = \frac{\pi}{2}$,we have $\phi = \frac{\pi}{2} - 	heta$. Thus,$\cos \phi = \sin 	heta$ and $\cot \phi = 	an 	heta$.Substituting these into (ii): $2x \sin 	heta + 3y 	an 	heta = 13$ (iii).To find the intersection $(\alpha, \beta)$,we solve the system of equations:$2x \cos 	heta + 3y \cot 	heta = 13$$2x \sin 	heta + 3y 	an 	heta = 13$Subtracting the equations or using Cramer's rule,we find the $y$-coordinate $\beta$.From $2x \cos 	heta + 3y \frac{\cos 	heta}{\sin 	heta} = 13$,we get $2x \cos 	heta \sin 	heta + 3y \cos 	heta = 13 \sin 	heta$.From $2x \sin 	heta + 3y \frac{\sin 	heta}{\cos 	heta} = 13$,we get $2x \sin 	heta \cos 	heta + 3y \sin 	heta = 13 \cos 	heta$.Subtracting the two resulting equations: $3y(\cos 	heta - \sin 	heta) = 13(\sin 	heta - \cos 	heta)$.Since $\sin 	heta 
eq \cos 	heta$ (for distinct points),we divide by $(\cos 	heta - \sin 	heta)$ to get $3y = -13$.Therefore,$\beta = -\frac{13}{3}$.

Let $A(2 \sec \theta, 3 \tan \theta)$ and $B(2 \sec \phi, 3 \tan \phi)$ where $\theta+\phi=\frac{\pi}{2}$,be two points on the hyperbola $\frac{x^2}{4}-\frac{y^2}{9}=1$. If $(\alpha, \beta)$ is the point of intersection of normals to the hyperbola at $A$ and $B$,then $\beta$ is equal to

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