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Area of the triangle formed by the lines $x -y = 0, x + y = 0$ and any tangent to the hyperbola $x^2 -y^2 = a^2$ is :-
$|a|$
$\frac{1}{2} |a|$
$a^2$
$\frac{1}{2} a^2$
Solution
Any tangent to the hyperbola at
$\mathrm{P}(\mathrm{a} \sec \theta, \mathrm{a} \tan \theta)$ is
$x \sec \theta-y \tan \theta=a$ …..$(1)$
Also; $\quad \mathrm{x}-\mathrm{y}=0$ ……$(2)$
$\mathrm{x}+\mathrm{y}=0$ …..$(3)$
from $(1),(2) $ and $(3) ;$ we get
$A\left(\frac{a}{\sec \theta-\tan \theta}, \frac{a}{\sec \theta-\tan \theta}\right) ; B\left(\frac{a}{\sec \theta+\tan \theta}, \frac{-a}{\sec \theta+\tan \theta}\right)$
${\&\,\, \mathrm{C}(0,0)} $
${\text { Area of } \Delta \mathrm{ABC}=\frac{1}{2}\left(\mathrm{x}_{1} \mathrm{y}_{2}-\mathrm{x}_{2} \mathrm{y}_{1}\right)}$
$={\frac{\mathrm{a}^{2}}{2}\left(\frac{-1}{\sec ^{2} \theta-\tan ^{2} \theta}-\frac{1}{\sec ^{2} \theta-\tan ^{2} \theta}\right)} $
${=\frac{\mathrm{a}^{2}}{2}(-2)=-\mathrm{a}^{2}=\mathrm{a}^{2}}$
