Answer
See explanations, $\frac{3}{4}$
Work Step by Step
Step 1. Given the equation $x=y^2$, differentiate both sides with respect to $x$: $1=2yy'$, which gives $y'=\frac{1}{2y}$,
Step 2. The slope of the line normal to the tangent line from point $(x_0,y_0)$ is $n=-\frac{1}{y'}=-2y_0$, and the line equation passing $(a,0)$ is $y=-2y_0(x-a)$
Step 3. For the above normal line to pass points on the parabola other than $(0,0)$, we have $y_0=-2y_0(x_0-a)$ which gives $a=x_0+\frac{1}{2}$. Since $x_0\gt0$, we have $a\gt \frac{1}{2}$
Step 4. For the other two normals to be perpendicular to each other, we have $n_1=-2y_1, n_2=-2y_2$ and $n_1n_2=-1$, which gives $4y_1y_2=-1$.
Step 5. Consider the symmetry of the curve. We have $y_2=-y_1$; using it in the above relation, we get $4y_1^2=1$. Thus we can obtain $y_1=\frac{1}{2}, y_2=-\frac{1}{2}$ and $x_1=x_2=\frac{1}{4}$, which gives $a=x_1+\frac{1}{2}=\frac{3}{4}$
