<p>If the three normals drawn to the parabola \(y^2 = 2x\) pass through the point \((a, 0)\) where \(a \neq 0\), then \('a'\) must be greater than</p>
Step-by-Step Solution
Key Concept: Three normals can be drawn to a parabola from a point on its axis only if that point lies beyond a critical distance equal to half the latus rectum from the vertex.
<p><strong>Step 1:</strong> For a parabola, at most three normals can be drawn from an external point on the axis.</p><p><strong>Step 2:</strong> For more than one normal to exist from a point on the axis, the point must lie beyond a critical distance from the vertex.</p><p><strong>Step 3:</strong> For the standard parabola $y^2 = 4ax$, three normals can be drawn from a point $(x_0, 0)$ on the axis if $x_0 > \frac{L}{2}$ where $L$ is the length of the latus rectum.</p><p><strong>Step 4:</strong> For the parabola $y^2 = 2x$, we have $4a = 2$, so $a = \frac{1}{2}$ and $L = 4a = 2$.</p><p><strong>Step 5:</strong> Therefore, $a > \frac{L}{2} = \frac{2}{2} = 1$.</p><p>∴ Answer is (d) $1$.</p>
Correct Answer: D