Given that, equation of the circle is x2 + y2 = 4.
The equation of the normal to the circle at (1, √3) is the equation of the line joining the points (1, √3) and (0,0).
⇒ y = √3 x (i)
∴ equation of normal is y = √3 x.
Now, the equation of the tangent to the circle at (1, √3) is
⇒ √3 y – 3 = -x +1
Putting y=0 , we get x=4
Hence, Δ AOB is formed by the positive x-axis and tangent and normal.
Now, Area of Δ AOB = Area of Δ AOC + Area of Δ ACB
= 2√3 sq. units
Hence, area of the triangle formed is 2√3 sq. units
Here, a =1 ,b=3 and f(x) = e2-3x+x2+1
∴ ⇒ hn=2
Rate this question :