Step-by-step Solution:

Any point on the hyperbola \( x^2-y^2=a^2 \) in parametric form is \( P(a\sec\theta, a\tan\theta) \).\nThe equation of the tangent at point \( P \) is:\n\[ x\sec\theta - y\tan\theta = a \]\n\nThe triangle is formed by this tangent and the lines \( y = x \) and \( y = -x \).\n\nFind the intersection of the tangent with \( y = x \):\n\[ x\sec\theta - x\tan\theta = a \implies x = \frac{a}{\sec\theta - \tan\theta} = a(\sec\theta + \tan\theta) \]\nThus, vertex \( A = (a(\sec\theta + \tan\theta), a(\sec\theta + \tan\theta)) \).\n\nFind the intersection of the tangent with \( y = -x \):\n\[ x\sec\theta - (-x)\tan\theta = a \implies x = \frac{a}{\sec\theta + \tan\theta} = a(\sec\theta - \tan\theta) \]\nThus, vertex \( B = (a(\sec\theta - \tan\theta), -a(\sec\theta - \tan\theta)) \).\n\nThe third vertex is the intersection of \( y=x \) and \( y=-x \), which is the origin \( O(0,0) \).\n\nThe area of the triangle with vertices \( (0,0) \), \( (x_1, y_1) \), and \( (x_2, y_2) \) is \( \frac{1}{2} |x_1 y_2 - x_2 y_1| \).\n\[ \text{Area} = \frac{1}{2} \left| a(\sec\theta + \tan\theta) \cdot (-a)(\sec\theta - \tan\theta) - a(\sec\theta - \tan\theta) \cdot a(\sec\theta + \tan\theta) \right| \]\n\[ = \frac{1}{2} \left| -a^2(\sec^2\theta - \tan^2\theta) - a^2(\sec^2\theta - \tan^2\theta) \right| \]\nSince \( \sec^2\theta - \tan^2\theta = 1 \):\n\[ \text{Area} = \frac{1}{2} | -a^2(1) - a^2(1) | = \frac{1}{2} | -2a^2 | = a^2 \]