Find the number of common normals to the curves $ x^2 + (y-1)^2 =1 $ and $y^2=4x$.
My take :
I formed a cubic in $m$ i.e. slope, so there'll be 3 normals. Please help.
Your first curve is a circle of radius 1, centered at $(0,1)$. Its normals are the lines through its center, that is, the lines $$y=mx+1$$ for arbitrary $m$ (this leaves out the vertical normal, but that's obviously not normal to the other curve). So now you just have to work out the values of $m$ for which the graph of $y=mx+1$ is normal to the graph of $y^2=4x$. Can you do that?
I guess not, so here goes.
From $y^2=4x$ we get $2yy'=4$, so $y'=2/y$. If $(a,b)$ is a point on the graph of $y^2=4x$, then