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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.

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Please explain how you formed a cubic in slope $m$. – hardmath Jun 30 '12 at 15:13
up vote 1 down vote accepted

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
1. the slope of the normal to the curve at that point is $-b/2$, and
2. $b^2=4a$.
So the equation of the normal is $$y-b=-(b/2)(x-a)$$ which we can write as $$y=-(b/2)x+(1/2)ab+b$$ But we want the normal to be $y=mx+1$, so $$(1/2)ab+b=1$$ Now combining that with $b^2=4a$, we get, after a little algebra, $$b^3+8b-8=0$$ and it's easy to show that equation has exactly one real zero, so there is exactly one common normal.

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so, no real values of m. Thus no common normals. – Bazinga Jul 1 '12 at 7:46
If you sketch the two curves, I think it's clear that there is at least one common normal. – Gerry Myerson Jul 1 '12 at 9:00

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