Question

The area of the triangle formed by the asymptotes and any tangent to the hyperbola $x^2-y^2=a^2$

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

Equation of any tangent to$x^2-y^2=a^2$
i.e. $\frac{x^2}{a^2}-\frac{y^2}{a^2}=1is\frac{x}{a}sec\theta -\frac{y}{a}tan\theta =1\to \left(1\right)$
$orxsec\theta -ytan\theta =a$
equation of other two sides of the triangle are
$x-y=0\dots \left(2\right)x+y=0\left(3\right)$
The two asymptotes of the hyperbola $x^2-y^2=a^2$
Are x-y=0 and x + y=0
Solving (1) (2) and (3) in pairs the coordinates of the vertices of the triangle are (0,0)
$\left(\frac{\alpha }{sec\theta +tan{\theta }^{\prime }}\frac{\alpha }{sec\theta +tan\theta }\right)And\left(\frac{\alpha }{sec\theta -tan{\theta }^{\prime }}\frac{-\alpha }{sec\theta -tan\theta }\right)-$
Area of triangle =$\frac{1}{2}|\frac{a^2}{se{c}^2\theta -ta{n}^2\theta }+\frac{a^2}{se{c}^2\theta -ta{n}^2\theta }|$=
$\frac{1}{2}(a^2+a^2)sincese{c}^2\theta -ta{n}^2\theta =1=a^2$