Area of triangle formed by any tangent at any point and asymptotes of the hyperbola xy = 2 is

The hyperbola given is xy = 2. The asymptotes of this hyperbola are y = x and y = -x.

Let's say the point on the hyperbola is (a, 2/a). The equation of the tangent at this point can be found using the property that the product of the perpendicular distances from the foci to any tangent of a hyperbola is equal to the square of the semi-transverse axis.

For the hyperbola xy = c^2, the semi-transverse axis is c, and the foci are (c, 0) and (-c, 0). So, the equation of the tangent is given by:

xx1 = c^2 => a*x = 2 => x = 2/a

The lines forming the triangle are y = x, y = -x, and x = 2/a.

The vertices of the triangle are the points of intersection of these lines, which are (2/a, 2/a), (-2/a, 2/a), and (0, 0).

The area of the triangle formed by these points can be found using the formula for the area of a triangle given the coordinates of its vertices:

Area = 1/2 * |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|

Substituting the coordinates of the vertices:

Area = 1/2 * |(2/a)(2/a - 2/a) + (-2/a)(2/a - 0) + 0(2/a - (-2/a))| = 1/2 * |0 + 4 - 0| = 1/2 * 4 = 2 square units

So, the area of the triangle formed by any tangent at any point and the asymptotes of the hyperbola xy = 2 is always 2 square units.