Find the area of the triangle formed by the positive x-axis, and both the normal and the tangent to the circle x^2 + y^2 = 4 at (x,y) = (1,sqrt(3)).
+129907 Slope of any tangent line to the circle = -x/y
At (1.sqrt 3) the equation of the tangent line is
y = (-1/sqrt (3) ) ( x -1) + sqrt (3)
This line intersects the x axis at
-sqrt (3)= (-1/sqrt (3)( x -1) mult through by sqrt (3)
3 = (x -1)
4 = x
The equation of the normal line at ( 1.sqrt 3) is
y = sqrt (3) ( x -1) + sqrt (3)
y = sqrt (3) x
This line intersects the x axis at x = 0
The base of the triangle formed = 4
The height of the triangle formed = sqrt (3)
Area = (1/2) (4) sqrt (3) = 2 sqrt (3)
Here's a pic :
