" The tangent at "(-1,2)" on "(y-1)^(2)=x^(2)(x+2)" meets the curve again at "P" .Then "P=

If the tangent at (1,1) on y^(2)=x(2-x)^(2) meets the curve again at P, then find coordinates of P.

(i) Find the point on the curve 9y^(2)=x^(3) where normal to the curve has non zero x -intercept and both the x intercept and y- intercept are equal. (ii) If the tangent at (1,1) on y^(2)=x(2-x)^(2) meets the curve again at P, then find coordinates of P (iii) The normal to the curve 5x^(5) -10x^(3) +x+2y+6=0 at the point P(0,-3) is tangent to the curve at some other point (s). Find those point(s)?

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

If the tangent at (1,1) on y^2=x(2-x)^2 meets the curve again atpis

If the tangent at P(1,1) on the curve y^(2)=x(2-x)^(2) meets the curve again at A , then the points A is of the form ((3a)/(b),(a)/(2b)) , where a^(2)+b^(2) is

If the tangent at P(1,1) on the curve y^(2)=x(2-x)^(2) meets the curve again at A , then the points A is of the form ((3a)/(b),(a)/(2b)) , where a^(2)+b^(2) is

If the tangent at (1,1) on y^2=x(2-x)^2 meets the curve again at P , then find coordinates of P .