Normal equation to the curve `y^(2)=x^3` at the point `(t^(2),t^(3))` on the curve is

The tangent to the curve y=x^(3) at the point P(t, t^(3)) cuts the curve again at point Q. Then, the coordinates of Q are

The equation of the normal to the curve y=x(2-x) at the point (2, 0) is

The equation of the normal to the curve y=e^(-2|x|) at the point where the curve cuts the line x = 1//2 is

The normal drawn on the curve xy=12 at point (3t,(4)/(t)) cut the curve again at a point having parameter point t, then prove that t_(1)=-(16)/(9t^(3))

Find the equation of tangent and normal to the curve y =3x^(2) -x +1 at the point (1,3) on it.

The equation of the normal to the curve y= e^(-2|x|) at the point where the curve cuts the line x=-(1)/(2), is

The equation of normal to the curve y=x^(3)-2x^(2)+4 at the point x=2 is -

Find the equation of normal to the curve y(x-2)(x-3)-x+7=0 at that point at which the curve meets X-axis.