1. Understanding the Problem and Key Concept: The Equation of a Normal

This problem asks us to determine the range of values for 'a' such that exactly three distinct normal lines can be drawn from the point $(a, 0)$ to the parabola $y^2 = 2x$.

The most effective way to approach this is by using the slope form of the equation of a normal to a parabola. For a standard parabola $y^2 = 4Ax$, a normal with slope $m$ has the equation: $y = mx - 2Am - Am^3$ Why this form? Each distinct real value of $m$ obtained from this equation corresponds to a unique normal line. If we substitute the coordinates of the point from which the normals are drawn, we get a cubic equation in $m$. The number of distinct real roots of this cubic equation will tell us the number of distinct normals that can be drawn from that point.

2. Identifying Parabola Parameters

First, we need to adapt the general equation of the normal to the specific parabola given: The given parabola is $y^2 = 2x$. We compare this to the standard form $y^2 = 4Ax$. By equating the coefficients of $x$: $4A = 2$ Solving for $A$, we find the parameter specific to this parabola: $A = \frac{2}{4} = \frac{1}{2}$ Why this step? The parameter $A$ is crucial as it defines the shape and scale of the parabola, and thus directly influences the equation of its normals.

3. Deriving the Normal Equation for the Specific Parabola

Now, substitute the value of $A = \frac{1}{2}$ into the general slope form of the normal equation: $y = mx - 2\left(\frac{1}{2}\right)m - \left(\frac{1}{2}\right)m^3$ Simplifying this expression, we obtain the equation of a normal to the parabola $y^2 = 2x$: $y = mx - m - \frac{1}{2}m^3$ Why this step? This specific equation is the tool we will use to find the slopes of normals passing through any given point.

4. Applying the Given Condition: Normals Pass Through $(a,0)$

We are given that three normals drawn to the parabola pass through the point $(a, 0)$. This means that the coordinates $(x, y) = (a, 0)$ must satisfy the equation of the normal derived in the previous step. Substitute $x=a$ and $y=0$ into the normal equation: $0 = m(a) - m - \frac{1}{2}m^3$ Why this step? By substituting the point, we transform the equation of a line (in terms of $x,y,m$) into an equation solely in terms of $m$ (for a given 'a'). The roots of this equation will be the slopes of the normals that pass through $(a,0)$.

To analyze the roots for $m$, let's rearrange this equation into a standard cubic polynomial form: $\frac{1}{2}m^3 + m - am = 0$ $\frac{1}{2}m^3 + (1 - a)m = 0$ Notice that $m$ is a common factor in both terms. We can factor it out: $m \left( \frac{1}{2}m^2 + (1 - a) \right) = 0$ Why this factoring? Factoring simplifies the cubic equation into a linear factor and a quadratic factor, making it easier to find its roots.

5. Analyzing for Three Distinct Normals

For three distinct normals to pass through the point $(a, 0)$, the cubic equation in $m$ must have three distinct real roots.

From the factored equation $m \left( \frac{1}{2}m^2 + 1 - a \right) = 0$, we can immediately identify one root: $m_1 = 0$ This root corresponds to the normal line $y = 0(x) - 0 - \frac{1}{2}(0)^3 \implies y=0$, which is the x-axis. This normal always passes through the vertex $(0,0)$ and is perpendicular to the axis of the parabola.

For the other two normals, we must analyze the quadratic factor: $\frac{1}{2}m^2 + 1 - a = 0$ To find the remaining roots, let's solve this quadratic equation for $m$: $\frac{1}{2}m^2 = a - 1$

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