Area of the triangle formed by the lines x - | vedclass

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(C) Any tangent to the hyperbola $x^2 - y^2 = a^2$ at point $P(a \sec 	heta, a 	an 	heta)$ is given by the equation:$x \sec 	heta - y 	an 	heta

Vedclass · Accepted Answer

$a^2$

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(C) Any tangent to the hyperbola $x^2 - y^2 = a^2$ at point $P(a \sec 	heta, a 	an 	heta)$ is given by the equation:$x \sec 	heta - y 	an 	heta = a$  --- $(1)$The other two lines are:$x - y = 0$  --- $(2)$$x + y = 0$  --- $(3)$Solving $(1)$ and $(2)$ simultaneously:$x(\sec 	heta - 	an 	heta) = a \implies x = a(\sec 	heta + 	an 	heta), y = a(\sec 	heta + 	an 	heta)$Solving $(1)$ and $(3)$ simultaneously:$x(\sec 	heta + 	an 	heta) = a \implies x = a(\sec 	heta - 	an 	heta), y = -a(\sec 	heta - 	an 	heta)$The intersection of $(2)$ and $(3)$ is the origin $(0, 0)$.The area of the triangle with vertices $(0, 0)$,$(x_1, y_1)$,and $(x_2, y_2)$ is $\frac{1}{2} |x_1 y_2 - x_2 y_1|$.Area $= \frac{1}{2} |a(\sec 	heta + 	an 	heta)(-a(\sec 	heta - 	an 	heta)) - (a(\sec 	heta - 	an 	heta))(a(\sec 	heta + 	an 	heta))|$Area $= \frac{1}{2} |-a^2(\sec^2 	heta - 	an^2 	heta) - a^2(\sec^2 	heta - 	an^2 	heta)|$Since $\sec^2 	heta - 	an^2 	heta = 1$,we get:Area $= \frac{1}{2} |-a^2 - a^2| = \frac{1}{2} |-2a^2| = a^2$.

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:

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