conic sections
The question is this:
For the parabola: $$ (x-1)^2 + (y-1)^2 = \left(\dfrac{x+y} {\sqrt2}\right)^2 $$
what is the condition on the point $(h,h)$ (which lies on the axis of
the parabola) that $3$ distinct normals can be drawn from it to the parabola?
Now, I have this doubt: from any point (inside the axis) on the axis we can draw three normals to the parabola (no concrete proof, just visualisation), right? So isn't this question wrong?
WLOG, we can consider the equation of the Parabola to be $\displaystyle y^2=4ax\ \ \ \ (1)$
From this, the eqaution of the normal at $P(am^2,-2am)$ is $y = mx – 2am – am^3\iff am^3+2a m+y-mx=0$
which is a Cubic Equation in $m$ having three roots $m_1,m_2,m_3$(say)
Using Vieta's formulas, $\displaystyle m_1+m_2+m_3=-\frac0a=0$
So, the ordinate of the centroid of the Triangle will be $\displaystyle\frac{-2a(m_1+m_2+m_3)}3=0$ hence, the centroid will lie on the $X$ axis which is evidently the axis of the Parabola $(1)$
Claude's technique can be extended slightly to find the the points and normal lines given a particular point off of the parabola. Using this, I obtained an animation of some of the normals.
Best Answer
After all the discussion in the comments part we have, perhaps, the solution to the edited question:
From a point on the parabola's axis it is possible to draw three different normals to the parabola iff the point is contained ("inside") the parabola but it is not the parabola's vertex (which could be considered not to be "inside" the parabola, anyway...but it's better, imo, to be thorough here and mention this).
Added: Thanks to one of the last comments of Henning below this answer, it is possible to see that the $\,y-$intercept of any normal to $\,y=x^2\,$ is $\,\frac12+a^2\;$ , from where it follows that no point on the parabola's axis and inside the parabola can have $\,y-$intercept $\,\le\frac12\;$ and have any hope of being one of the wanted points from which three normals to the parabola can be drawn!