If the tangent at P(1,1) on y²=x(2-x)² meets the curve again at Q , then Q is

Step-by-step solution

array l y²=x(2-x)² y²=x³-4 x²+4 x (i) 2 y dy dx =3 x²-8 x+4 dy dx = 3 x²-8 x+4 2 y [ dy dx ]_ P = 3-8+4 2 =- 1 2 array y²=x(2-x)² y²=x³-4 x²+4 x (i) 2 y dy dx =3 x²-8 x+4 dy dx = 3 x²-8 x+4 2 y [ dy dx ]_ P = 3-8+4 2 =- 1 2 Equation of tangent at P is: y-1=- 1 2 (x-1) x+2 y-3=0 x+2 y-3=0 Using y= 3-x 2 in (i), we get: ( 3-x 2 )² array l =x³-4 x²+4 x 4 x³-17 x²+22 x-9=0 ...(ii) array which has two roots 1,1 (Because of (ii) being tangent at (1,1) ). Sum of 3 roots = 17 4 array l 3 rd root = 17 4 -2= 9 4 Then, y= 3-