What is the area of the triangle formed by th | vedclass

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(A) The equation of any tangent to the hyperbola $x^{2} - y^{2} = a^{2}$ at point $P(a \sec 	heta, a 	an 	heta)$ is given by $x \sec 	heta - y 	a

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$a^{2}$

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(A) The equation of any tangent to the hyperbola $x^{2} - y^{2} = a^{2}$ at point $P(a \sec 	heta, a 	an 	heta)$ is given by $x \sec 	heta - y 	an 	heta = a$ ......$(i)$The given lines are $x - y = 0$ ......(ii) and $x + y = 0$ ......(iii).Solving these lines in pairs,we find the vertices of the triangle as:$A = \left( \frac{a}{\sec 	heta - 	an 	heta}, \frac{a}{\sec 	heta - 	an 	heta} ight)$,$B = \left( \frac{a}{\sec 	heta + 	an 	heta}, -\frac{a}{\sec 	heta + 	an 	heta} ight)$,and $C = (0, 0)$.Since one vertex is the origin $(0, 0)$,the area of the triangle $ABC$ is given by $\frac{1}{2} |x_{1}y_{2} - x_{2}y_{1}|$.Area $= \frac{1}{2} \left| \left( \frac{a}{\sec 	heta - 	an 	heta} ight) \left( -\frac{a}{\sec 	heta + 	an 	heta} ight) - \left( \frac{a}{\sec 	heta + 	an 	heta} ight) \left( \frac{a}{\sec 	heta - 	an 	heta} ight) ight|$Area $= \frac{a^{2}}{2} \left| \frac{-1}{\sec^{2} 	heta - 	an^{2} 	heta} - \frac{1}{\sec^{2} 	heta - 	an^{2} 	heta} ight|$Since $\sec^{2} 	heta - 	an^{2} 	heta = 1$,we have:Area $= \frac{a^{2}}{2} |-1 - 1| = \frac{a^{2}}{2} |-2| = a^{2}$.

What is the area of the triangle formed by the lines $x - y = 0$,$x + y = 0$,and any tangent to the hyperbola $x^{2} - y^{2} = a^{2}$?

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