Animated Solution for Mathematics - Conic Sections: If the three normals drawn to the parabola, $y^2=2x$ pass through the point $(a,0)a\ne 0$, then 'a' must be greater than :

The parabola is defined by the equation $y^2=2x$. Comparing this to the standard form $y^2=4Ax$, we identify the parameter $4A=2$, which yields:

This parameter $A$ is the fundamental constant that dictates the geometry of the parabola. We will utilize this value to derive the properties of the normals.

To describe any normal to the parabola, we employ the slope-form equation:

Substituting our specific value $A=\frac{1}{2}$ into this equation, we obtain the specific normal equation for our parabola:

We are interested in normals passing through the point $(a,0)$. Since this point lies on the line, its coordinates must satisfy the normal equation. Substituting $x=a$ and $y=0$ into the equation, we get:

To solve for the slopes $m$, we factor the expression:

The factored equation reveals that either $m=0$ or $a-1-\frac{m^2}{2}=0$. The case $m=0$ corresponds to the normal lying along the x-axis.

For the remaining two normals to be real and distinct, the quadratic part must yield two distinct real roots:

For $m$ to be real and non-zero, the right-hand side must be strictly positive. This requires:

If $a>1$, we obtain two distinct non-zero values for $m$, which, combined with the $m=0$ case, provide exactly three distinct real normals.

In the general case for a parabola $y^2=4Ax$, the condition for three distinct normals passing through $(a,0)$ is $a>2A$. Since $2A=1$ in our specific case, the condition $a>1$ is confirmed as the threshold for the existence of three distinct normals.