If the three normals drawn to the parabola, y² = 2x pass through the point (a, 0) a ≠ 0, then 'a' must be greater than :

Solution

Maths sir · 5 steps

Write the normal in slope form, put (a, 0) into it, and ask when you get three real slopes.

1. Match the standard form. Compare y² = 2x with y² = 4Ax. This gives 4A = 2, so A = 1/2.
2. Normal in slope form. The normal with slope m to y² = 4Ax is y = mx - 2Am - Am³. With A = 1/2 it becomes y = mx - m - m³/2.
3. Pass through (a, 0). Put x = a, y = 0. This gives 0 = ma - m - m³/2, which factorises as below.

   m(a - 1 - m²/2) = 0

4. Count the slopes. One root is always m = 0, which is the axis itself. The other two come from m² = 2(a - 1). These are real and distinct, and different from 0, only when a - 1 > 0.
5. So. Three normals pass through (a, 0) only when a > 1, which is option (b).

The idea to keep. For y² = 4Ax, three normals pass through (h, 0) only when h > 2A. Put the point into the slope-form normal and make the quadratic in m² give a positive value.

Answer from the NTA final answer key. Solution by PrepWiser.