Find the set of points on the axis of the parabola $y^2-2y-4x+5=0$ from which all three normals to the parabola are real.

Question

$$(y-1)^2=4(\frac 14)(x-1)$$

Equation of normal to the tangent in the terms of slopes

$$y-1=m(x-1)—2am-am^3$$

Since normal is from the axis, y=1

$$0=m(x-1)-\frac m2 -\frac{m^3}{4}$$ $$m^3+m(6-4x)=0$$

Since all m are real $$m_1^2+m_2^2+m_3^2\ge 0$$ $$(m_1+m_2+m_3)^2-2(m_1m_2+m_2m_3+m_3m_1)\ge0$$ $$0-2(6-4x)\ge 0$$ $$x\ge \frac 32$$

So the set is ${(x,1):x\ge \frac32}$

But the answer is $x\ge 3$

What is going wrong?

Answer 1

We have

$y^2-2y+1-4x+4=0$

$(y-1)^2-4(x-1)=0$

$(y-1)^2=4(x-1),\color{blue}{a=1}$. Not $a=1/4$.

With the proper value of $a$ your calculation for $x$ should be correct.