BITSAT Mathematics Question (2021) — Solution

Step-by-step solution

array l y^ 2 =x(2-x)^ 2 y^ 2 =x^ 3 -4 x^ 2 +4 x (i) \\ 2 y dy dx =3 x^ 2 -8 x+4 \\ dy dx = 3 x^ 2 -8 x+4 2 y \\ [ dy dx ]_ P = 3-8+4 2 =- 1 2 array y^ 2 =x(2-x)^ 2 y^ 2 =x^ 3 -4 x^ 2 +4 x (i) 2 y dy dx =3 x^ 2 -8 x+4 dy dx = 3 x^ 2 -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 =x^ 3 -4 x^ 2 +4 x \\ 4 x^ 3 -17 x^ 2 +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- 9 4 2 = 3 8 \\ Q is ( 9 4 , 3 8 ) array