\[\begin{array}{1 1}(1)\frac{\pi}{4}&(2)\frac{\pi}{3}\\(3)\frac{\pi}{6}&(4)\frac{\pi}{2}\end{array}\]

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$\large\frac{x^2}{25} +\frac{y^2}{9} $$=1$-----(1)

$\large\frac{x^2}{8} -\frac{y^1}{8} $$=1$-----(2)

$\large\frac{x^2}{8} + \frac{y^2}{8} $$=1$

$a= \large\frac{1}{25} $

$b= \large\frac{1}{9} $

$A= \large\frac{1}{8} $

$B= \large\frac{-1}{8} $

$\large\frac{1}{a} -\frac{1}{b}= \frac{1}{1/25}-\frac{1}{1/9}$$=25-9=16$------(3)

$\large\frac{1}{A}-\frac{1}{B}=\large\frac{1}{1/8} = \frac{1}{1/-8}$$=8+8=16$------(4)

From (3) and (4) the given two curves intersect orthogonality

$=> \large\frac{\pi}{2}$

Hence 4 is the correct answer.

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