To find the area of the triangle, we must analyze the intersections of the given lines and the tangent line to the hyperbola.
1. Identify the asymptotes:
The given lines x−y=0 (which is y=x) and x+y=0 (which is y=−x) are the asymptotes of the hyperbola x2−y2=a2. They intersect at the origin (0,0).
2. Equation of the tangent:
For a hyperbola x2−y2=a2, the tangent at any point (x0,y0) on the hyperbola is given by xx0−yy0=a2. We can represent the point of tangency using parametric coordinates as P(asecθ,atanθ). The equation of the tangent at this point is:
xsecθ−ytanθ=a
3. Find the vertices of the triangle:
The vertices of the triangle are the origin (0,0) and the intersection points of the tangent with the two asymptotes.
Intersection with y=x:
Substitute y=x into the tangent equation:
xsecθ−xtanθ=a
x(secθ−tanθ)=a
Since secθ−tanθ1=secθ+tanθ, we get x=a(secθ+tanθ).
Thus, the first intersection point is (a(secθ+tanθ),a(secθ+tanθ)).
Intersection with y=−x:
Substitute y=−x into the tangent equation:
xsecθ+xtanθ=a
x(secθ+tanθ)=a
So, x=a(secθ−tanθ).
Thus, the second intersection point is (a(secθ−tanθ),−a(secθ−tanθ)).
4. Calculate the Area:
The area of a triangle with vertices (0,0), (x1,y1), and (x2,y2) is given by 21∣x1y2−x2y1∣.
Using our coordinates:
x1=a(secθ+tanθ)
y1=a(secθ+tanθ)
x2=a(secθ−tanθ)
y2=−a(secθ−tanθ)
Area=21∣(a(secθ+tanθ)⋅−a(secθ−tanθ))−(a(secθ−tanθ)⋅a(secθ+tanθ))∣
Using the identity sec2θ−tan2θ=1:
Area=21∣−a2(1)−a2(1)∣
Area=21∣−2a2∣=a2
Correct Option: 2. a2
