<s>[INST] 3 normals to the parabola 
Suppose there exists three normal lines from the point $(r, 0)$ to the
  parabola $x = y^2$, and $r >  \frac{1}{2}$, one of which is the
  x axis. Determine the value of r where the two other normal lines would be
  perpendicular?

I found the $\frac{dy}{dx} (parabola)= \frac{1}{2\sqrt{x}} $ .
Thus gradient of normal, is $-2\sqrt{x}$, and the equation of normal is $y=-2\sqrt{x} (x-r) $.
If I equate the equation of normal with the parabola, 
I get $ \pm \sqrt{x}=-2\sqrt{x} (x-r) $, 
which solving will result in $x=\pm\frac{1}{2} +r$, 
however I am stuck and I cannot find an exact value of r. Where did I go wrong and how should I have approached this question?
 [/INST] In the diagram below, if the two normals are to be perpendicular, then we have a right triangle whose vertices are the three marked points, so that
$$(2a)^2 = 2((a-0)^2+(a^2-r)^2) = 2 a^2+2 a^4-4 a^2 r+2 r^2.$$
Also, since the lines are normal to the parabola, we see that the slope of the upper normal, which is $-2a$, must be equal to the slope of the line, so that
$$-2a = \frac{a}{a^2-r}.$$
Solving these two equations for $a$ and $r$ gives $r = \frac{3}{4}$ and $a = \pm\frac{1}{2}$.

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