Find the area of triangle formed by the positive x axis and the tangent and normal to the curve x^2 +y^2=9 at (2,root 5)?

x² + y² = 9
(d/dx)(x² + y²) = (d/x)9
2x + 2y(dy/dx) = 0
2y(dy/dx) = -2x
dy/dx = -x/y

Slope of the tangent at (2, √5)
= -2/√5

Slope of the normal at (2, √5)
= -1 / (-2/√5)
= √5/2

Let a be the x-intercept of the tangent at (2, √5).

Slope of the tangent at (2, √5):
(0 - √5) / (a - 2) = -2/√5
(√5) / (a - 2) = 2/√5
2a - 4 = 5
2a = 9
a = 9/2
Hence, the tangent at (2, √5) cut the x-axis at (9/2, 0)

Let b be the x-intercept of the normal at (2, √5).

Slope of the normal at (2, √5):
(b - √5) / (0 - 2) = √5/2
(b - √5) / (-2) = √5/2
b - √5 = -√5
b = 0

Hence, the normal at (2, √5) cut the x-axis at (0, 0)

Area of the triangle formed by the positive x-axis and the tangent and normal at (2, √5)
= (1/2) × (9/2) × √5
= (9√5)/4 (square units)