Show that area of the traingle formed by the tangent and the normal at the point(a,a) on the curve y²(2a-x)=x³ and the line x=2a is 5a²/4 sq. units.

Equation of the given curve is y ² (2a – x) = x ³

y ² (2a – x) = x ³

Differentiating both sides w.r.t x, we get

Slope of tangent at 

∴ Slope of normal at 

Equation of tangent at (a, a) is

(y – a) = 2 (x – a)

∴ y = 2x – 2a + a = 2x – a ...(1)

Equation of normal at (a, a) is

∴ 2y – 2a = – x + a

Equation of the given line is

x = 2a ...(3)

Solving (1) and (3), we have

x = 2a and y = 2 (2a) – a = 4a – a = 3a

Solving (2) and (3), we have

Solving (1) and (2), we have

x = a and y = a

 be the vertices of the triangle.

Are of ∆ABC

Thus, the area of triangle formed is  square units.