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?
Best Answer
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 $r=\frac{3}{4}$