If three distinct normal can be drawn to the parabola $y^2-2y=4x-9$ from $(2a,b)$. then range of $b$ is
solution i try
$y^2-2y+1=4x-9+1$
$(y-1)^2=4(x-2)$ parabola of vertex is at $(1,2)$ and focus is $(3,2)$
equation of normal to the parabola is $y-1=m(x-2)-2m-m^3$
$y=mx-4m-m^3+1$
Normal through $(2a,b)$ is $b=2am-m^3+1$ How to find range of $b$
Comparing the equation of this parabola with the standard form, we can see that a suitable parametrization is $$x=t^2+2$$ $$y=2t+1$$
In which case the equation of the normal is $$y-(2t+1)=-t(x-t^2-2)$$
Substituting $(x,y)=(2a,b)$ leads to the cubic equation $$f(t)=t^3+t(4-2a)+1-b=0$$
In order for the cubic equation $y^3+Py+Q=0$ to have three distinct real roots, we require $P<0\implies a>2$ and also that $4P^3+27Q^2<0$
So you can get a range of values for $b$ in terms of $a$