3 normals to the parabola

Question

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?

Answer 1

The tangents and the normals form a square

The slope of a tangent is $m=\pm \frac{1}{2\sqrt{x}}$

Which must be $\pm 1$

$-\frac{1}{2\sqrt{x}}=1\to x=-\frac{1}{4}$

The tangent has equation $y=x+\frac14$

the intersection with the parabola is

$$x=\left(x+\frac14\right)^2 \to x=\frac14;\;y=\frac{1}{2}$$

Normals have slope $m^{\perp}=\pm 2\sqrt{x}$

The normal has equation

$$y-\frac{1}{2}=-(x-\frac14)$$

the intersection with $x$ axis is $r=\frac{3}{4}$

Answer 2

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 $a = \pm\frac{1}{2}$.